Mezo Playing Zoma
Today, Mezo is playing a game. Zoma, a character in that game, is initially at position $$$x = 0$$$. Mezo starts sending $$$n$$$ commands to Zoma. There are two possible commands:  'L' (Left) sets the position $$$x: =x - 1$$$;  'R' (Right) sets the position $$$x: =x + 1$$$. Unfortunately, Mezo's controller malfunctions sometimes. Some commands are sent successfully and some are ignored. If the command is ignored then the position $$$x$$$ doesn't change and Mezo simply proceeds to the next command.For example, if Mezo sends commands "LRLR", then here are some possible outcomes (underlined commands are sent successfully):   "LRLR" — Zoma moves to the left, to the right, to the left again and to the right for the final time, ending up at position $$$0$$$;  "LRLR" — Zoma recieves no commands, doesn't move at all and ends up at position $$$0$$$ as well;  "LRLR" — Zoma moves to the left, then to the left again and ends up in position $$$-2$$$. Mezo doesn't know which commands will be sent successfully beforehand. Thus, he wants to know how many different positions may Zoma end up at.
In the example, Zoma may end up anywhere between $$$-2$$$ and $$$2$$$.
The first line contains $$$n$$$ $$$(1 \le n \le 10^5)$$$ — the number of commands Mezo sends.The second line contains a string $$$s$$$ of $$$n$$$ commands, each either 'L' (Left) or 'R' (Right).
Print one integer — the number of different positions Zoma may end up at.
print(int(input()) + 1)
n = int(input())
s = input()
print(n + 1)
import sys
sys.setrecursionlimit(100000000)
input = lambda : sys.stdin.readline().strip()
write = lambda x: sys.stdout.write(str(x) + '\n')
n = int(input())
s = input()
print(n + 1)