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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main {
	public static void main(String[] args)throws IOException {
		FastReader sc = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);
		int x = sc.nextInt();
		int count = 0;
		int tmp = x;
		x = 0;
		while (x != 360) {
			x = x + tmp;
			if (x > 360) {
				x = x - 360;
			}
			//System.out.println(x);
			count++;
		}
		System.out.println(count);





	}
}





class FastReader {
	BufferedReader br;
	StringTokenizer st;

	public FastReader() {
		br = new BufferedReader(new
		                        InputStreamReader(System.in));
	}

	String next() {
		while (st == null || !st.hasMoreElements()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException  e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	int nextInt() {
		return Integer.parseInt(next());
	}

	long nextLong() {
		return Long.parseLong(next());
	}

	double nextDouble() {
		return Double.parseDouble(next());
	}

	String nextLine() {
		String str = "";
		try {
			str = br.readLine();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return str;
	}
}