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) {
		// TODO Auto-generated method stub
		Scanner scn = new Scanner(System.in);
		int X = scn.nextInt();
      	int p = 360 / X;
      if(p * X == 360){
      	System.out.println(p);
      }
      else{
        System.out.println(p + 1);
      }
	}
}