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)

#include<bits/stdc++.h>

using namespace std;

#define pii pair<int,int> 
#define F first 
#define S second
#define int long long 
#define mpr make_pair
typedef long long ll;

const int maxn = 2e5+10;
const int mod = 1e9+7;
const ll inf = 1e18+10;

int n, m, k;
int p[maxn];
vector<int> g[maxn];

signed main()
{
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

   cin>> n;
   cout<< ((n*360)/(__gcd(n,360ll)))/ n;
}