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: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
    int am = 1;
    int count = 0;
    int r = 360;
    Scanner sc = new Scanner(System.in);
    int x = sc.nextInt();
    sc.close();

        if(r%x == 0){
            System.out.println(r/x);
        }else{
            while(am != 0){
                am = r%x;
                count += r/x;
                r = r+am;
                if(am == 0){
                    System.out.println(count);
                }
            }
        }
    }
}