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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>

using namespace std;

int gcd(int x, int y)
{
  if (y == 0) return x;
  return gcd(y, x % y);
}

int main()
{
  int X;
  cin >> X;
  
  int lcm = X * 360 / gcd(X, 360);
  cout << lcm / X << endl;
  
  return 0;
}