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: Runtime Error)

package main;
import java.util.Scanner;
public class Main {
	int d;
	public void solve() {
	Scanner sc = new Scanner(System.in);
	int a = sc.nextInt();
	int s=a;
	d=1;
while(true) {
	a+=s;
	 d=d+1;
	// System.out.println(d);
	 if(a==180) {
		 d=d*2;
		 break;
	 }else if(a>360) {
		 d=d*2;
		d= d-1;
		 break;
	 }
}

System.out.println(d);
}


	public static void main(String[] args) {
		new Main().solve();
	
	}
}