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){
Scanner sc=new Scanner(System.in);
  int x=sc.nextInt();
  if(360%x==0){
System.out.print(360/x);
  }else{
    int i=1;
while(true){
int y=x*i;
if(y%360==0){
System.out.print(i);
  break;
}
i++;
}

  }
}
}