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: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int x = sc.nextInt();
        
        for(int i=1; i<=180; i++){
            if(((x*i)>=360) &&((x*i)%180 == 0) ){
                System.out.println(i);
                break;
            }
        }
        
        
    }
}