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)

#include<bits/stdc++.h>
using namespace std;
using lli = long long int;
using ld = long double;
using ulli = unsigned long long int;
#define filein freopen ("in.txt", "r", stdin)
#define fileout freopen ("out.txt", "w", stdout)
#define dbg(args...) do {cerr << #args << " : "; debug(args); } while(0)
#define all(v) ((v).begin()),((v).end())
#define size(x) (int) x.size() 
#define pb emplace_back
#define mp make_pair
#define PI 3.14159265
#define mod 1000000007
void debug () {cerr << endl;}
template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }
template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }
template < typename T, typename ... hello>void debug( T arg, const hello &... rest) {cerr << arg << ' ';debug(rest...);}
int main(){
	ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    for(int t = 1; t <= T; t++){
		int x;
		cin >> x;
		cout << 360 / x;
	}
}