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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define MOD 1000000007
using Graph = vector<vector<int>>;
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
namespace mp = boost::multiprecision;
// 任意長整数型
using Bint = mp::cpp_int;
// 仮数部が1024ビットの浮動小数点数型(TLEしたら小さくする)
using Real = mp::number<mp::cpp_dec_float<1024>>;
  long long as[200010];

int main (){
  int N;
  if(360%N==0)
  cout<<360/N;
  else
  cout<<N;
}