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

Solution in C (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void){

    int x = 0;
    int k = 0;
    scanf("%d", &x);

    if(360%x == 0){
        k = 360 / x;
    } else {
        for(int i = 2; (360 * i % x) == 0; i++){
            if(i == x){
                k = x * x;
            } else {
                k = 360 * i / x;
            }
        }
    }

    printf("%d", k);
    return 0;
}