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: Accepted)

import java.util.*;
public class Main {
    public static void main(String[] args){
      // 入力
      Scanner sc = new Scanner(System.in);
      int X = sc.nextInt();
      
      // 算出
      int K = 1;
      int deg = X;
      while(deg % 360 != 0){
        deg += X;
        K++;
      }
      
      // 出力
      System.out.println(K);
    }

}