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<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<math.h>
#include<map>
#include<iomanip>
#include<queue>

using namespace std;

int main(){
	int n;
	cin >> n;
	int ans;
	if(360%n==0){
		ans=360/n;
	}else{
		if(360%(180-n)==0){
			ans=360/(180-n);
		}else{
			ans=360;
		}
	}
	cout << ans << endl;
}