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>
#include<vector>
#include<cmath>
#include<math.h>
#include<cstdlib>
#include<assert.h>
#include<string>
#include<sstream>
#include<algorithm>
#include<ctime>
#include<iostream>
#define ll long long 
#define ld long double
#define pb push_back
#define rep(i,n) for(i=0;i<n;i++)
#define rep1(j,n) for (j=0;j<=n;j++)
#define PI 3.14159265358979323846
#define mod 1000000007
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL) 
using namespace std;
/*void input(vector<ld>& v,ld n)
{
	ld i,io;
	rep(i,n)
	{
		cin>>io;
		v.pb(io);
	}
}*/
int main()
{
	fastio;
	int n;
	cin>>n;
	cout<<(360)/n<<endl;	
}