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.*;
import java.io.*;

public class Main{
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.valueOf(br.readLine());
		if(n<=90)
		System.out.println(360/n);
		else
			System.out.println(360/(180-n));
	}
}