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 void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        long n = sc.nextInt();
        long tmp = lcm(360, n);
        long tmp2 = tmp / n;
        
        // System.out.println(tmp);
        System.out.println(tmp2);
    }
    
    public static long lcm(long m, long n) {
        return m * (n / gcd(m, n));
    }
    
    public static long gcd(long a, long b) {
        if (b == 0) return a;
        
        return gcd(b, a % b);
    }
}