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.io.PrintWriter;
import java.util.Scanner;

public class Main {
    int gcd(int x, int y) {
        return y == 0 ? x : gcd(y, x % y);
    }
    void solve(Scanner sc, PrintWriter pw) {
        int x = sc.nextInt();
        int g = gcd(360, x);
        pw.println((360 / g));
    }
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        new Main().solve(sc, pw);
        pw.flush();
        pw.close();
        sc.close();
    }
}