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 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
      
      	int x = sc.nextInt();
      
      	int a = 1;
      
      	int b = 0;
      	int k = 0;
        while(a != 0){
            k += 360;
        	a = k % x;
        }
          	b = k / x;
      
      	System.out.print(b);
    }
}