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>
using namespace std;

#define FAST ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define MAXX 1000005


#define PI 3.14159265358979323846264338327950

#define ll  long long int
ll mod = 1000000007;
ll dp[MAXX];


int main()
{
	FAST;
	ll n, t , h = 1;
	cin >> n;
	t = 360;
	h = (360*n)/(__gcd(n, t));
	h /= n;
	cout << h ;
    return 0;
	
}