Timetable
There are two bus stops denoted A and B, and there $$$n$$$ buses that go from A to B every day. The shortest path from A to B takes $$$t$$$ units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as $$$a_1 < a_2 < \ldots < a_n$$$ for stop A and as $$$b_1 < b_2 < \ldots < b_n$$$ for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least $$$t$$$ units of time later than departs.It is known that for an order to be valid the latest possible arrival for the bus that departs at $$$a_i$$$ is $$$b_{x_i}$$$, i.e. $$$x_i$$$-th in the timetable. In other words, for each $$$i$$$ there exists such a valid order of arrivals that the bus departed $$$i$$$-th arrives $$$x_i$$$-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the $$$i$$$-th departed bus arrives $$$(x_i + 1)$$$-th.Formally, let's call a permutation $$$p_1, p_2, \ldots, p_n$$$ valid, if $$$b_{p_i} \ge a_i + t$$$ for all $$$i$$$. Then $$$x_i$$$ is the maximum value of $$$p_i$$$ among all valid permutations.You are given the sequences $$$a_1, a_2, \ldots, a_n$$$ and $$$x_1, x_2, \ldots, x_n$$$, but not the arrival timetable. Find out any suitable timetable for stop B $$$b_1, b_2, \ldots, b_n$$$ or determine that there is no such timetable.
Consider the first example and the timetable $$$b_1, b_2, \ldots, b_n$$$ from the output.To get $$$x_1 = 2$$$ the buses can arrive in the order $$$(2, 1, 3)$$$. To get $$$x_2 = 2$$$ and $$$x_3 = 3$$$ the buses can arrive in the order $$$(1, 2, 3)$$$. $$$x_1$$$ is not $$$3$$$, because the permutations $$$(3, 1, 2)$$$ and $$$(3, 2, 1)$$$ (all in which the $$$1$$$-st bus arrives $$$3$$$-rd) are not valid (sube buses arrive too early), $$$x_2$$$ is not $$$3$$$ because of similar reasons.
The first line of the input contains two integers $$$n$$$ and $$$t$$$ ($$$1 \leq n \leq 200\,000$$$, $$$1 \leq t \leq 10^{18}$$$) — the number of buses in timetable for and the minimum possible travel time from stop A to stop B.The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_1 < a_2 < \ldots < a_n \leq 10^{18}$$$), defining the moments of time when the buses leave stop A.The third line contains $$$n$$$ integers $$$x_1, x_2, \ldots, x_n$$$ ($$$1 \leq x_i \leq n$$$), the $$$i$$$-th of them stands for the maximum possible timetable position, at which the $$$i$$$-th bus leaving stop A can arrive at stop B. 
If a solution exists, print "Yes" (without quotes) in the first line of the output.In the second line print $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ ($$$1 \leq b_1 < b_2 < \ldots < b_n \leq 3 \cdot 10^{18}$$$). We can show that if there exists any solution, there exists a solution that satisfies such constraints on $$$b_i$$$. If there are multiple valid answers you can print any of them.If there is no valid timetable, print "No" (without quotes) in the only line of the output.
def read():
    return list(map(int, input().split(' ')))

def fail():
    print('No')
    exit(0)
(n, t) = read()
aa = read()
xx = read()
res = [0] * n
prevX = 0
prevV = -10
for i in range(n):
    x = xx[i] - 1
    if x < prevX or x < i:
        fail()
    prevX = x
    res[i] = prevV = max(aa[i + 1] + t if x > i else aa[i] + t, prevV + 1)
for i in range(n):
    x = xx[i] - 1
    if x + 1 < n and res[x] >= aa[x + 1] + t:
        fail()
print('Yes')
print(*res)
(n, t) = map(int, input().split(' '))
aa = list(map(int, input().split(' ')))
xx = list(map(int, input().split(' ')))
res = [0] * n
prevX = 0
prevV = -10
for (i, (a, x)) in enumerate(zip(aa, xx)):
    x -= 1
    if x < prevX or x < i:
        print('No')
        exit(0)
    curV = max(aa[i + 1] + t if x > i else aa[i] + t, prevV + 1)
    res[i] = curV
    prevX = x
    prevV = curV
for (i, (a, x)) in enumerate(zip(aa, xx)):
    x -= 1
    if x + 1 < n:
        if res[x] >= aa[x + 1] + t:
            print('No')
            exit(0)
print('Yes')
print(*res)
def read():
    return list(map(int, input().split()))

def fail():
    print('No')
    exit(0)
(n, t) = read()
aa = read()
xx = read()
bb = [0] * n
prevX = 0
prevV = -10
for i in range(n):
    x = xx[i] - 1
    if x < prevX or x < i:
        fail()
    prevX = x
    bb[i] = prevV = max(aa[i + 1] + t if x > i else aa[i] + t, prevV + 1)
for i in range(n):
    x = xx[i] - 1
    if x < n - 1 and bb[x] >= aa[x + 1] + t:
        fail()
print('Yes')
print(*bb)