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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Runtime.InteropServices;
using System.Text;

namespace AC
{
    class Program
    {
        static void Main(string[] args)
        {
            var X = ReadInt();
            var ans = 1;
            var angle = 360 - X;
            while (angle != 0)
            {
                ans++;
                angle -= X;
                if (angle < 0) angle = 360 - angle;
            }

            Console.WriteLine(ans);
            Console.ReadLine();
        }

        static string Read() { return Console.ReadLine(); }
        static int ReadInt() { return int.Parse(Read()); }
        static long ReadLong() { return long.Parse(Read()); }
        static string[] ReadArrayString() { return Read().Split(' '); }
        static int[] ReadArrayInt() { return Read().Split(' ').Select(s => int.Parse(s)).ToArray(); }
        static long[] ReadArrayLong() { return Read().Split(' ').Select(s => long.Parse(s)).ToArray(); }
    }
}