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 <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <tuple>
#include <vector>
using namespace std;
typedef long long ll;
ll const INF = 1LL << 60;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

int main() {
  double X;
  cin >> X;
  cout << (360 / gcd(X, 360LL)) << endl;
  return 0;
}