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)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ll long long
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const int INF = 1000000000;

int main() {    
    int X;
    cin >> X;
    X = min(X, 180-X);
    
    for (int k=1; true; k++){
        if ((k*X)%360 == 0){
            cout << k << endl;
            return 0;

        }
    }

}