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 <iostream>
#include <cstdio>
#include <cassert>
#include <ctime>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

#define all(x) (x).begin(), (x).end()

#ifdef LOCAL
#define dbg(x) cerr << #x " = " << x << endl;
#define idbg(x) #x " = " << x
#include "pretty_print.h"
#else
#define dbg(x)
#define idbg(x) ""
#endif

#ifdef _WIN32
#  define LLD "%I64d"
#else
#  define LLD "%lld"
#endif

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;

template <typename T> T sqr(T x) { return x * x; }
template <typename T> T abs(T x) { return x < 0? -x : x; }
template <typename T> T gcd(T a, T b) { return b? gcd(b, a % b) : a; }
template <typename T> bool chmin(T &x, const T& y) { if (x > y) { x = y; return true; } return false; }
template <typename T> bool chmax(T &x, const T& y) { if (x < y) { x = y; return true; } return false; }


int main(int /* argc */, char** /* argv */)
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    #ifdef LOCAL
    assert(freopen("inp", "r", stdin));
    assert(freopen("out", "w", stdout));
    #endif
    int x;
    while (cin >> x) {
        cout << 360 / gcd(360, x) << endl;
    }

    cerr << "Time execute: " << clock() / (double)CLOCKS_PER_SEC << " sec" << endl;
    return 0;
}