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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
public class Main {
    public static long gcd(long x,long y){
        if(x < y) return gcd(y, x);
        if(y == 0) return x;
        return gcd(y, x % y);
    }
    public static long lcm(long x,long y){                                                              
        return x*y/gcd(x,y);                                                                                                                                                                                                                
    }
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();		
		System.out.println(360/gcd(360,a));
	}
}