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: Accepted)

using System;
using static System.Console;

namespace A
{
    internal class Program
    {
        public static void Main(string[] args)
        {
            var angle = int.Parse(ReadLine());
            WriteLine(Solve(angle));
        }

        private static double Eps = .000001;

        static int Solve(int rot)
        {
            (double, double y, double currentDegree) pos = (0, 0, rot);
            var p = Move(pos, rot);
            var k = 1;

            while (!(Math.Abs(p.x) <= Eps && Math.Abs(p.y) <= Eps))
            {
                p = Move(p, rot);
                k++;
            }

            return k;
        }

        static (double x, double y, double currentDegree) Move((double x, double y, double currentDegree) p, int rot)
        {
            var movedPos = (p.x, p.y, p.currentDegree);
            var rad = Math.PI * p.currentDegree / 180.0;
            movedPos.x += Math.Cos(rad);
            movedPos.y += Math.Sin(rad);
            movedPos.currentDegree += rot;
            return movedPos;
        }
    }
}