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) {
        Scanner sc = new Scanner(System.in);
        long x = sc.nextLong();
        if (360 % x == 0) {
            System.out.println(360 / x);
        } else {
            long lcm = lcm(x, 360);
            System.out.println(lcm / x);
        }
    }

    private static long gcd(long a, long b) {
        long r;
        while( (r = a % b) != 0 ) {
            a = b;
            b = r;
        }
        return b;
    }

    private static long lcm(long a, long b) {
        return a * b / gcd(a,b);
    }
}