Problem:


Score : 200 points

Problem Statement

Takahashi is standing on a two-dimensional plane, facing north. Find the minimum positive integer K such that Takahashi will be at the starting position again after he does the following action K times:

Go one meter in the direction he is facing. Then, turn X degrees counter-clockwise.

Constraints

1 \leq X \leq 179

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the number of times Takahashi will do the action before he is at the starting position again.

Sample Input 1

90

Sample Output 1

4

Takahashi's path is a square.

Sample Input 2

1

Sample Output 2

360



Sample input:
90


Sample output:
4

Submission status: Runtime Error

Solution code:
def main():
    X = int(input())

    for i in range(10**5):
        t += X
        if t % 360 == 0:
            return i + 1

print(main())

Submission status: Runtime Error

Solution code:
import math, sys
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from copy import deepcopy
from functools import lru_cache
from heapq import heapify, heappop, heappush
from itertools import accumulate, combinations, permutations
input = sys.stdin.readline
mod = 10**9 + 7
ns = lambda: input().strip()
ni = lambda: int(input().strip())
nm = lambda: map(int, input().split())
nl = lambda: list(map(int, input().split()))

def main():
    x = ni()

    print(360//x))


if __name__ == '__main__':
    main()

Submission status: Runtime Error

Solution code:
print(360 // input())