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: Accepted)

/* A - Takahashikun, The Strider */

#include <bits/stdc++.h>
using namespace std;

int gcd(int a, int b){
  int r;
  r = a % b;
  if(r == 0) return b;
  else return gcd(b, r);
}

int main(){
  int X, G;
  cin >> X;
  G = gcd(X, 360);
  cout << 360 / G << endl;
}