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 (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdbool.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define rep(i,n) for(int i=1; i <(n); i++)
int main(){
  int x,k;
  scanf("%d",&x);
  if(360%x==0){
    printf("%d\n",360/x);
  }else if(360%(180-x)==0){
    k=180-x;
    printf("%d\n",360/k);
  }else{
    printf("%d\n",360/(360-x));
  }
  return 0;
}