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, start, end) for (long long i = start; i < end; ++i)
#define repreverse(i, start, end) for (long long i = start; i >= end; --i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define len(x) ((long long)(x).size())
#define lcm(a, b) ((a) / __gcd((a), (b)) * (b))
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vllvll = vector<vll>;
using vc = vector<char>;
using pll = pair<ll, ll>;
template<class T>void print1d(T x,ll n=-1){if(n==-1)n=x.size();rep(i,0,n){cout<<x[i]<<' ';}cout<<'\n';}
template<class T>void print2d(T x,ll r=-1,ll c=-1){if(r==-1)r=x.size();if(c==-1)c=x[0].size();rep(i,0,r)print1d(x[i],c);}
template<class T, class U>bool haskey(T mp, U key) { return mp.find(key) != mp.end(); }
template<class T, class U>bool isin(T el, U container) { return find(all(container), el) != container.end(); }
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
template<class T>bool even(T n) { return ! (n & 1); }
template<class T>bool odd(T n) { return n & 1; }
template<class T>ld deg2rad(T deg) { return M_PI * deg / 180.0; }
template<class T>ld rad2deg(T rad) { return 180.0 * rad / M_PI; }
ll intpow(ll a,ll n){ll p=1;while(n){if(n&1)p*=a;a*=a;n>>=1;}return p;} 
const long double pi = M_PI;
const long long big = 1LL << 50;
const long long inf = 1LL << 60;
const long long mod = 1e9 + 7;



int main()
{
        ll X;
        cin >> X;
        ll now = X;
        rep(i, 1, inf) {
                if (now % 360 == 0) {
                        cout << i << endl;
                        return 0;
                }
                now += X;
        }
}