Okay, welcome back everybody. We are here for our developer sync. We've been out in Hawaii for a while and we got some really great work done. We're gonna resume our daily stand-ups and here we go. So let's start with Gebs. Hey, so Pencore was not ready and ready as I hoped. So they're approaching from two directions. So one is using a full 2024 image consistent with what we were using before. But it's a much bigger image so it's, you know, anytime there's an update, it could be up to like two gigabytes and every user has to download. So that is clearly not ideal and that's not the direction they want to go. But I've asked them to get us that just for the starting point so we can, you know, get moving on that. Yeah, the issue that they're having is with a conflict in Alpine Linux with libcrypto and QT 5.0. So we're solving that in different ways of dealing with that, connecting them with Alpine and we'll see if we can just actually just resolve the issue. Another option is that we bump the QT version to 5.5.14 and there's some detail around how the implications of that and how we do that and all that sort of stuff, but it's probably more than we can kind of deal with now. Yeah, just a quick question. None of the choices they've taken are to use our existing recipes that build a working QT image. Well, that's the one thing you wanted. That's the one thing you wanted for a long time. Are you saying you're getting a buy-in? I remitted a QT, yeah. I can't remember how many it is, but it's, you know, a number of gigabytes, yeah. Well, I mean, I think it's wonderful they could shave our image size down, but you might think that for the expense of a gig and a half of this, the storage, that we can afford it and they could move forward with whatever works and then they could work. The impression I got was it's not a storage issue, it's a download issue. It makes it a great internet connection.  Yeah, especially if it's an update every single time. But let's, we can discuss the soft line. Try to stick to our updates. All right, Derek, how's it going? Okay, okay. So I'm gonna see what's there. I was talking about getting the SJ230B2 update, which is compatible with the new SJ2211 that can just, so the new design is a little bit different size and the power comes out differently. So we needed to update the laser grid design and the 3D printer design. So I think I'd promised Josh that I'd not done one same. I did not get that done. So I did get that done today. So if you have 20 minutes or so, Josh, put it on the laser and cut it out for me. And then I've been printing parts for the updated version as well. So as soon as I get the board from Kevin, I can throw that in the new design and test anything up. So that's been going on. And then tomorrow, I would like to be sending one of these to our, hopefully our assembler for, at least for the initiative kits, but hopefully for everything. So I just got off an email to hand coordinate that. So hopefully I'll get those off tomorrow. With the caveat that it won't have the latest board in it, but I think that's okay. We can get it off the new board. So yeah, that's been. All right, what's coming up for tomorrow? So I have all of the parts for SJ240 printed except for the top and the back. So I do have a couple of weeks that can finish those. So I'll probably work it on that. In the morning, and then I think I, we need to thank Josh and Johnny and whoever else has raised their hand on that binding parts team. I need to stick on that a little bit and try to talk about some stuff. So maybe try and do that tomorrow. All right. Yeah. Try to give them a little heads up. Yeah. I'll sing up some today so you can try this. Yeah. Okay, sounds great. I'll just throw myself on the spot here. Because why not? So I had promised last week that I'd be working on the roadmap and trying to consolidate a lot of the documentation that we went through in Hawaii and created in Hawaii. When I dug into that, it was a little bit more extensive than I thought it was going to be. We've got something like 40 spec documents going back a couple of years. And then there's another, doesn't earn more, excuse me, roadmaps that I need to consolidate as well. But, so I don't have all of that work done, but I will at least get a roadmap for the roadmap out today, which I think will still be useful. And you guys can all tell me what you think of it later on. So, yeah, so that's what I've been up to. But I did enjoy my holiday, which I didn't know that I needed, but I did. So, Ken. You learned something new every day. I did not realize that our update process was us pushing a complete new unit out to every device on an update. That's interesting and carries its own ramifications of discussion from a serious point. Sorry, Ken, to interrupt. I don't think that's actually the case. I think they attempt to push small binary diffs, but given the size of the image it sounds like, it would end up being a lot of diffs anyway. So, but let's not get sidetracked on that. What are you working on? So anyway, yeah, I'm working on, actually I'm going through some of the things that Jarvis and Kevin went through. I'm almost completed with the refactoring and removal of the mark two skill, the second mark two skill to be in the tab and pulling that into the actual date, score, and closure for client and enclosure proper. So I'm knee deep in that, trying to get messages from both the viewing message bus and the regular message bus and keeping the existing volume skill in sync. So that when you were to push a button and it changes the volume, it wouldn't confuse the volume skill. So I throw a map, I anticipate having a full request ready to go tomorrow, so that it will find its way into the next kind of full release. And now I will move over to using Chris's API and getting back into the incremental machine learning project. Okay. All right, thanks. Are you still using the mark two feature branch? Is that we're still doing this work? Yeah, in other words, I haven't committed yet. That's where the full request is going to be put. There will be a few changes, lines of code that I'll ask you to commit on the whole blue system skill, just to dampen the actual output of the IPC command. So just two comments, comment two lines out and then just make sure that that's released in all future skills. Yeah, but other than that, I'll be on the market right here. Okay. Chris there, how was your Thanksgiving? It was okay, a little lonely.  Made a lot of food on the Thanksgiving dinner. But you have like seven dogs, right? So. That's true, I could share with them. But I did get a version of the query I was struggling with working. And I have the tagger itself working again, basically to the point where our initial iteration is supposed to be. So. Economic committed right now. I think what I'm going to do tomorrow is to test the whole path.  You know, run from the batch jobs, make sure the whole thing kind of works end to end. And then once that's done, I will, I mean, we'll have to work with Ken on some of the, some of the whiteboard stuff, but because I'm curious as to what, what do you want me to do next? And then you send an email out about, you know, maybe building a whiteboard contributor tool. To help us get further faster with training the whiteboard. But then there's also some stuff that can take over possibly. And now that's a big stuff goes. So I just wasn't sure what my priority would be after I get this initial iteration of the tagger. Kind of ready to go. Okay. So it sounds like you've got another day of debugging and testing ahead of you. So let's figure that out between now and the end of tomorrow then. Yeah, I don't know. I circle up with the Ken and the guys and see if there's anything that you can help out with or that they need to help with on that front. Otherwise we'll, we'll get back into the job board and there's no shortage of stuff to work on. I mean, Chris, I think if you're ready, we can meet briefly tomorrow. You will be ready to make sure that everything's working there. You know, go over that. If nothing else, that'll validate. Maybe open up some stuff and you're ready for that. Whenever you're ready. Okay. All right, great. And Josh, any updates for us? I and I have been working through the, we'll couple things. NASA, we'll find out tomorrow whether we were awarded the following contract. And Johnny and I have been working on a financial model, which is what I'm looking at right now. That runs out through the end of 2022. And it allows us to deliver everything that we promised. And then I'm patiently waiting for an operating system for this thing. So when under that stunt, that would be great. Let me know if you want me to have a discussion with our vendor. Our vendor? Panticle. Oh. I mean, we're not, there was a lot of unknowns going into this, but as Gev said earlier, we're close. I've also been given a lot of thought to messaging, but that doesn't really belong at this meeting. So I will politely step aside. And then whenever Derek gets me stuff, I will start printing and what you're cutting and so on. All right, fair enough. Okay, and out of left field comes Emily. Emily, do you have an update for us? We, Kevin had talked with Emily in Hawaii about qualifying some PCBA companies for us. Just checking to see if... Oh, sorry. How's that going? I'm gonna need to offer him. I set up some logins for some of the companies, but he had to get some more information together for me to be able to do more. Okay, so you're still waiting on some info from Kevin. Okay, feel free to ping him. Again, he just got back recently from Hawaii. So, but he's back and he's working today. Okay, we'll do. All right, awesome. Okay, well, thanks everybody. That'll be it for today. And...


Hi, I'm Zor. Welcome to Unizor Education. Well, it's time to talk about properties of the function limits. Well, these properties are very much analogous to properties of limits of the sequences, and I will definitely rely on that material. But at the same time, we have two different definitions of the function limits, one of them through sequences and another through epsilon-delta language. So I will try to prove whatever the properties we will discuss using both ways. Okay, so limits of the functions have certain properties. This is a lecture which is part of the advanced course for high school students and teenagers of mathematics presented on Unizor.com. I suggest you to watch this lecture from the website because the website has very detailed notes for each lecture. Plus, you have the ability to take exams, which is like a very good self-testing kind of thing. The site is free, so properties of the limits. As I said, properties of the function limits are very much analogous to properties of the sequence limits. And the first one is if my function converges to some number a as x converges to some number r, then function which is a times f of x converges to a times a as x converges to r. Now, let's consider that a is positive. If a is negative, it's exactly the same, so it's just easier to consider a as a positive thing. Now, I will explain it the first time in a little bit more details about how to prove this using the sequences. But for all other properties, it's exactly the same, so I might actually be much more brief in this case. So, the first definition of the function limit deals with any sequence of arguments which is converging to r should result in sequence of the function of these arguments converging to a. That's what it means from the definition number one of the function limit. This is supposed to be for any sequence of arguments which is converging to r. Okay, fine. So, what do we have to prove using this definition if we want to prove this? Well, we want to prove that if you take any sequence of arguments which is converging to r, then sequence of a times f of x n converges to a times a. Now, since we know that for any sequence of arguments converging to r, if you have a sequence converging to a from the properties which have been proven for the sequence limits, if you have a sequence which is converging to some number, a times this sequence converging to a times that number. So, that's basically, this theorem is basically direct consequence from the corresponding theorem for limit of the sequences. And the property is correspondingly related to the properties of the sequence limits. Okay, so that's according to my definition of the limit of the function through sequences. I spend a little bit more time maybe this first time because all other will be exactly the same for all other properties. I'll just refer to the corresponding property of the sequences. But how about the second definition? That's more important and I think it's just more interesting as a nice logical exercise. So, what if I would like to prove this using epsilon delta definition of the limit? So, let's remind that this thing that the function has limit a as x converges to r means that for any epsilon greater than 0 exists such delta that as long as my x is within delta neighborhood of its limit point from this immediately follows that f at x would be within epsilon neighborhood of limit a. That's what's given. Now, what do I have to prove in this case? I have to prove that for any epsilon greater than 0 exists delta such that as long as my x is within neighborhood within delta neighborhood of its limit point r my function would be within epsilon neighborhood from a times a. This is needed to be proven. So, this is given and this is to be proven. Okay, so for any epsilon. Alright, so let's just take any epsilon first. Take any epsilon. Now, how can I find this delta? Well, in this case it's really very very simple because if I will choose delta equals to epsilon divided by a we have agreed that a is positive, right? So, let's take this. Take any epsilon and then, or I'm sorry, it's not delta, it's just epsilon prime. Now, for this epsilon prime number two, we find delta such that using this one. So, instead of epsilon, we will use epsilon prime. So, for this epsilon prime there is a delta such that this implies this epsilon prime. Now, what is epsilon prime? Again, this is epsilon divided by a, right? So, it's f of x minus a less than epsilon divided by a. So, now we have proven this. Well, multiplied by a, a is positive, so it will be here, a, a and not this one. And we have proven this particular inequality which we need, right? So, again, how did we prove this thing? Well, we actually physically found that delta using what? First, from our epsilon, which is completely anything whatever we want, we have to prove it for any epsilon. We have derived this new epsilon prime and then using this instead of epsilon, epsilon prime. We know basically that this is given. And that's how we got this and from this we got this. That's it, very simple. Next. Next property is, so the first one is function times constant converges to its limit times this constant. Now, the second property is limit of sum of two functions, each of them converging to something as x converging to r. Then I am stating that sum is converging to sum of the limits. And I will prove exactly the same way I actually built for any epsilon I will build delta. All right, by the way, as far as definition using the sequences, again, it's obvious because this means if I would choose any sequence of x, x1, x2, xn, etc., which is converging to r, then this would be right. So for any sequence of x, xn, which is converging to r, this would be this, which is a sum of two converging sequences. And we have already proven the theorem for sequences. So the sequence of the sum converges to the limit, which is sum of limits of each one of them. So with sequences, it's very simple. Now with epsilon delta, that's a little bit different, but as simple actually. Let's just think about it. Let's just fix any delta, any epsilon first. What we have to prove is, well, first what's given. What's given is that exists such delta that as long as x is within delta neighborhood of r, my f at x would be within delta neighborhood of a, epsilon neighborhood of a. Let's put it delta 1. Why? Because for the same epsilon, we have delta 2 for g. So as long as we know some kind of an epsilon, using the convergence of x, of f at x to a, we have found delta 1. As long as x within delta 1 neighborhood of r, this is true. And then we found, again, using the convergence of the g, we found delta 2. As long as x within delta 2 neighborhood of r, this is correct. Well, how about if I will take delta equal to minimum of delta 1 and delta 2. So if this means that there is such a delta 1 neighborhood, this is delta 2 neighborhood. And what is delta neighborhood? Well, if this is r, now this might be my delta 1 neighborhood, r plus delta 1. And this is r minus delta 1. Now, this might be r minus delta 2 and r plus delta 2. But if delta is minimum of them, then it's this interval, which is smaller than any of these. And since it's smaller, then both conditions are true. So basically, we have found such delta for any epsilon when both conditions are true. Well, if both conditions are true, then obviously we can write the following. f at x plus g at x minus a plus b, right? We have to prove that for any epsilon, for any epsilon, there is such a delta that this is less than of epsilon if my x minus r within delta neighborhood. How can I find this delta? Okay, here is a very simple solution. I will use an axiom, which everybody knows, right? So absolute value of sound is less than equal than absolute value of... Basically, if a and b both are positive, then this is equality. If a and b are negative, this is also equality. But if they are a different sign, like minus 2 and plus 5, now this is 3, but this is 2 and this is 5, it would be greater. So in this case, I will do the following. I will do f at x minus a plus g of x minus b, right? Now, I can make for any delta, for any epsilon, I can find such delta that both of them are true, right? So let me just choose, for this epsilon, I will choose epsilon divided by 2 and find such a delta that will be delta 3 already, right? That if x minus r is less than delta 3 by absolute value, then this is true. And for the same epsilon 2, I will find delta 4 such that if x is within delta 4 neighborhood, this is true, that this is less than epsilon divided by 2. And then I will take, again, the minimum of delta 3 and delta 4 and that would be my final delta when both of them will be true. So this is less than epsilon over 2 and this is less than epsilon over 2, which is epsilon. So for any epsilon, I will find such a delta when this is true. So first, from epsilon, I go to epsilon divided by 2. From epsilon divided by 2, I find this one, this delta 3, so that this is less than epsilon divided by 2 if x is in delta 3 neighborhood. Then for the same epsilon divided by 2, I find delta 4 neighborhood of r as long as x is in there, this is true. And that's where the whole thing is true. So I can always use all these manipulations to find for any epsilon, I can find delta. The third property is product, as you probably have guessed already. Product is just a little bit trickier. So this is given as x converges to r. What I have to prove is this. Again, speaking about definition of the product in terms of sequences, it has already been proven for sequences. So if I take any sequence of arguments which converges to r, any, then the corresponding sequence of function f is converging to a, corresponding sequence of g of xn converges to b, and the product of two converging sequences has a limit which is a product of two limits. So with sequences, it's done. Now, epsilon delta is more interesting, I would say. Well, let's do some analysis. f of xg of x minus ab. That's what we have to somehow prove that this is a very, very small, any kind of degree of smallness, infinitesimal actually, if x is converging to r. So we have to prove that this is infinitesimal as x converging to r. How can I prove it? Here's a little trick. I subtract it and add it a times gx. Well, how did I come up with this? Difficult to say because probably some very, very long time ago somebody showed it to me and since I remember this trick. But anyway, it is a little trick and why did I do it? Here is why. Because now I can separate f and g. Here is how. Remember this? So this is my a and this is my b. So it's less than or equal to f of x times g of x minus a times g of x, which is this. Right? I took g of x factor out. Plus, and here I will factor out absolute value of a. All right? Now, why did I do it? Because here this is really infinitesimal and this is infinitesimal as x goes to r. This is given. Now, this is the constant. So infinitesimal times constant is infinitesimal, right? And the only thing which is kind of strange thing is I don't know what to do with this one. Right? Well, here is what I can suggest. G goes to b. Right? So let's just assume that b is not equal to zero. Because if b is equal to zero, it's slightly different but very simple as well. Now, if b is not equal to zero, then g, since g is converging to b, eventually it will be, let's say, in this interval. Right? Well, not exactly. It would be b plus b over two and b two, which is b minus b over two. Right? That's what I can definitely say. So if g is converging to b, then it will be greater than b minus b two and less than b plus b two. Eventually. So it will be in the b two neighborhood. Why? Well, that's quite obvious since I know that for any epsilon greater than zero exists such delta that as long as x minus r is less than or equal to delta, it immediately follows that g of x minus b would be less than epsilon. Right? So choose epsilon equal to b over two. It's for any epsilon. Right? So I choose this one. And now I have this. This is equivalent of this. Now, why did I do it? Well, primarily for this particular reason. Since I know that g of x right now is some kind of a value in between these two as long as my delta is sufficiently small. I can actually say that with sufficiently small, with x sufficiently close to r, with x sufficiently close to r, this is less than or equal to, let's say, f of x minus a times b plus b two. I just replace, which is 3b over two. Now, instead of b over two, I can choose anything actually here. I can choose one, for instance. For epsilon equal to one, I can always find sufficiently close x to r that it would be between b minus one and b plus one. Now, if I choose, instead of this b one, b plus one, it will be b plus one. So any kind of a b plus something would work here. Right? And the sign is less than or equal plus a g of x minus b. Right? Now, I have constant here and constant here. And knowing this, now this is analysis. Now, how can I actually find out exactly whatever I need, which is for any epsilon, find delta such and such that this is less than delta? Well, very simple. If I choose any epsilon, then we will find a different epsilon one, let's say equals to epsilon divided by two b plus one. Something like this. That would be even better. Right? Now, whenever my x is sufficiently close to r, I can find such delta one that f of x minus a would be less than this. But since we are multiplying by b plus one, as a result, I will have less than epsilon over two. Right? So there is such a delta one when this is less than this. On the other hand, there is for the same, let's say let's take epsilon two equals to epsilon divided by absolute value of a. I will find delta two such that if x is sufficiently close, closer than delta two to b to r, then this would be smaller than this. And multiply by a. Yeah, I should put two here as well. And also epsilon over two. So this will be epsilon over two less than or equal. And this will be less than epsilon over two. So to make both of them satisfied, they just have to take minimum between delta one and delta two. When both of them are chosen. Again, for any epsilon, I choose epsilon one and find from convergence of f of x to a, find delta one when this is smaller than this. Then I choose using convergence of g of x to b, I choose for epsilon two equals to this one delta two such that this is true. And then minimum between delta one and delta two is the delta which I need because then this is less than epsilon over two. And this is epsilon over two. So the sum would be equal to epsilon whatever I need. So what's difficult about this? Just this trick. Everything else is just trivial inequalities. And the last property. Now the last property is, now we have multiplied, now we have to divide, right? Now I'm not going to divide two functions. I will just have function this. So if f of x goes to a, I need one over f of x goes to one over a. Obviously a not equal to zero as x converges to r. Now, again, speaking about sequences, it's obvious it just follows from the property for the sequence limit. Now talking about epsilon delta language, so what we have to find out is this thing an infinitesimal if x converges to r. But let's think about this is equal to, right? Well a minus f of x or f of x minus a, the same thing because it's an absolute value. If I will prove that this is just some kind of a limited value bounded on both sides, then I'm fine. As long as this is not, because this is infinitesimal, right? As x goes to r. So all we need is to prove that this is not infinitesimal as well. Because if it's bounded, if it's far from zero, then infinitesimal divided by this would be infinitesimal. How can I prove it? Well, very simply, again, let's have epsilon equals to a over two. Now, I find, obviously I can find delta one such that if x is closer than delta one, f at x minus a is closer than n a over two. Now, what is this? It means that a minus a over two, f at x, a plus a over two, which is a over two, and this is three a over two, right? So that's how we bound it. As long as f at x converges to something which is not equal to zero, eventually it will be close enough to separate it from zero, right? So if this is zero and this is a, obviously if it converges to a, after a certain point when argument is significantly close to r, I will be separated from zero, a over two and three a over two. Or I can choose a over four, it doesn't really matter. As long as this is not equal to zero, I'm far from zero. Now, what does it mean? It means that for this particular delta one, as long as my x is within delta one neighborhood, this thing, how can I increase the fraction? I will decrease the denominator. So I will take the smaller one, right? I will take the left, left border, the smaller one. A over two, within delta one neighborhood. A over two is smaller than epsilon square over f at x. Now, now I have to prove that this is infinitesimal, but this is the constant, right? So it's quite simply to prove that this is infinitesimal, because for any epsilon I can find delta such that this is, I will choose epsilon, any epsilon, right? Then I will choose epsilon prime equals to epsilon divided by a square over two, right? No, the other way around. Multiply it. Now, for this epsilon, epsilon prime, I can find such delta that this thing is less than this. And then if I divide it by this, I will have, so if this is less than epsilon, epsilon, then f at x minus a divided by a is less than epsilon. So I need this delta. Now, combined with the previous, now this is not just the first delta which we were using. Before that, we have found delta such that f at x minus a was basically, that f at x was close enough to a, so it's separated from zero, right? So this is separated from zero. So now I have one delta and that delta, so I'll probably take the smallest, the minimum of these two deltas, and then I will get the minimum delta is the one where both. This is true and this is true. And that's why the whole thing would be less than epsilon, whatever I'm supposed to prove. So in this and all other cases, it's a little bit maybe tricky to find the proper delta. You have to really separate, separately satisfy a few different conditions. But if you want to satisfy different conditions and different conditions require different deltas, you just take the minimum of these deltas. So whenever x is closer and closer and closer than any other requirement, then all the requirements will be satisfied. My f at x would be greater than a over two. And in the previous cases, we have similar conditions, etc.  And it's a little bit more like a textbook. I'm not taking any shortcuts. I'm really proving something. Okay, this is delta one. This is delta two. This is delta three. And we're taking minimum of all of them, etc. I don't want to go to all these scruples details during the lecture because my purpose was for you to understand really how this thing is supposed to be proven. So it's divided into smaller pieces. Each small has its own requirement for the delta. And then we take the minimum of these delta to satisfy all the requirements. Okay, so basically that's it. I strongly suggest you to read the notes for this lecture. Go to unizord.com to derivative and function limits. This is lecture called properties of the limits. And read it very thoroughly. It's relatively easy reading, but a little bit more, I would say, mathematically rigorous than whatever I was just trying to explain. All right. Thank you very much and good luck.


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Denver Police. I'm Andrew Richmond. Sir, just step back over here. An instructor here at the Denver Police Academy. Go ahead. Get down and drop. And today we're in the Scenario House. I ain't going anywhere. We'll take you outside. My Scenario House is a structure that we built, designed to allow us to do scenarios and role-playing for our recruits here at the Academy. The act of being able to physically go through these steps of being a police officer and interact with role-players.  Then do it. It gives them a more realistic understanding of how it's going to be when it's on the street. What are you guys doing? In today's scenario, two people are arguing in an apartment. Step back in here. And that's all they know. And in this scenario in particular, we put a gun on one of them. Gun! I actually think it's one of the most essential pieces of our training. Our arrest control, our law, as well as the report writing. This is where we get to see them put all of those things together. March 7.


Once again, this historic moment, first time that two women have done a spacewalk together. Christine Cook there, the first to come out of the quest airlock. People ask me a lot, were you scared when you did your spacewalk? And I really just talk about re-channeling fear into focus. And I think that I first learned that and discovered that that was something I could do at Science and Math. You were so busy all the time there. You know, I remember looking at the amount of homework and thinking how is this going to get done in one night. I just lacked, I guess, the confidence coming into the situation to realize it. But to have people that sat me down and pushed me harder than I would have pushed myself was really instrumental. My whole world opened up to me. You know, I was from Jacksonville, North Carolina, kind of a small town, small city. And then I went to NCSSM where there were people from Raleigh and Charlotte and all these big cities that I had heard about. And they had big opinions. They had seen so much more than I had seen. And just to realize that, you know, there was so much more to explore, so much more to learn. And the amount that my mind opened during that time is just, can't be understated. Walking down the hallway after the downlink event, one of the young women here who's a student just stopped me to say, this was so inspiring to be an astronaut and to set the record for the longest time in space by a woman. I think she truly embodies that there are no ceilings. It really helps the current students think about, well, that could be me. If you want to reach for the stars, you know, you can get there. And it's been just a really wonderful, very long-distance relationship. Hi, Christina. My name is Jocelyn, and I'm a student here at the North Carolina School of Science and Mathematics. And I just want to congratulate you on your 300th day in space. Go Unis! I want to thank you for dedicating yourself to being a role model for women instead.  It's so uplifting to know that someone who walked these very halls has accomplished something so great for the world. Thank you for inspiring all of us here at NCSSM to keep reaching for the stars. Still seeing Christina Cook there on the end of the robotic arm. It is going to be moving her to the external logistics carrier number one. One saying that I say to people that helped me a lot when I was sort of at a crossroads and figuring out what to do next, which is the case for a lot of students at Science and Math. Do what scares you. Oftentimes the things that we feel like are just outside of our reach, they're piquing our interest, we're passionate about it, but we might not actually think that we can do it. You just reach just outside of what you think might be possible. When you do achieve that, it pays dividends in terms of your fulfillment, how rewarding it is, and also how much you've given back to the world in achieving that.