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;

int gcd(int a, int b) 
{ 
    // Everything divides 0  
    if (a == 0) 
       return b; 
    if (b == 0) 
       return a; 
   
    // base case 
    if (a == b) 
        return a; 
   
    // a is greater 
    if (a > b) 
        return gcd(a-b, b); 
    return gcd(a, b-a); 
} 
   

int main()
{
	
	
	int x;
	cin>> x;
	

	int k = 360/(gcd(360,x));
	cout<< k;

	return 0;
	
}