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>
#define rep(i, n) for (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define all(x) x.begin(),x.end()
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }

ll x;
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin>>x;
    int k=1;
    while(x*k%360!=0)k++;
    cout<<k<<endl;
    //cout<<solve()<<endl;
    return 0;
}