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)

// AGC46-A
#include <bits/stdc++.h>
#define endl "\n"
#define rep(i,n) for(int i=0;i<int(n);i++)
#define e_b emplace_back
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> ipair;
typedef pair<ll,ll> lpair;
template <class T>ostream &operator<<(ostream &o,const vector<T>&v) //vectorの中身を見る
{o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")<<v[i];o<<"}";return o;}
void map_p(map<int,int>& d){cout<<"map: ";for(auto a : d){cout<<"{"<<a.first<<":"<<a.second<<"}";}cout<<endl;} //mapの中身を見る
void set_p(set<int>& d){cout<<"set: ";for(int a : d){cout<<a<<" ";}cout<<endl;} //setの中身を見る


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(12);

    int x;
    cin>>x;
    cout<<360/x<<endl;
    return 0;
}