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: Accepted)

#include <algorithm>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <memory>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <assert.h>

using namespace std;

typedef long long ll;

#define sz size()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(c) (c).begin(), (c).end()
#define rep(i,a,b) for(ll i=(a);i<(b);++i)
#define per(i,a,b) for(ll i=b-1LL;i>=(a);--i)
#define clr(a, b) memset((a), (b) ,sizeof(a))
#define ctos(c) string(1,c)
#define print(x) cout<<#x<<" = "<<x<<endl;

#define MOD 1000000007

long long gcd(long long a, long long b){
  return b == 0LL ? a : gcd(b, a % b);
}

long long lcm(long long a, long long b){
  return a / gcd(a, b) * b;
}

int main(){
  ll x;
  cin>>x;
  ll a = lcm(360,x);
  cout << a/x << endl;
  return 0;
}