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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
#define CLR(t,v) memset(t,(v),sizeof(t))
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>void chmin(T&a,const T&b){if(a>b)a=b;}
template<class T>void chmax(T&a,const T&b){if(a<b)a=b;}


ll nextLong() { ll x; scanf("%lld", &x); return x;}


using D = long double;
const D PI = acos(-1.0);
const D EPS = 1e-10;
struct P {
  D x, y;
  P(D x=0, D y=0) : x(x), y(y) {}

  P& operator+=(const P& o) { x += o.x; y += o.y; return *this; }
  P& operator-=(const P& o) { x -= o.x; y -= o.y; return *this; }
  P& operator*=(const P& o) { return *this = {x*o.x - y*o.y, x*o.y + y*o.x}; }
  P& operator*=(const D& r) { x *= r; y *= r; return *this; }
  P& operator/=(const D& r) { x /= r; y /= r; return *this; }
  P operator-() const { return {-x, -y}; }

  D norm() const { return x*x + y*y; }
  D abs() const { return sqrt(norm()); }
  D arg() const { return atan2(y, x); }
  bool isZero() const { return std::abs(x) < EPS && std::abs(y) < EPS; }
  /** 象限 */
  int orth() const { return y >= 0 ? (x >= 0 ? 1 : 2) : (x < 0 ? 3 : 4); }
  static P polar(const D& rho, const D& theta = 0) { return {rho * cos(theta), rho * sin(theta)}; }
};
std::ostream &operator<<(std::ostream &os, P const &p) { return os << "(" << p.x << ", " << p.y << ")"; }
std::istream &operator>>(std::istream &is, P &p) { D a, b; is >> a >> b; p = P(a, b); return is; }
P operator+(const P& p, const P& q) { return P(p) += q; }
P operator-(const P& p, const P& q) { return P(p) -= q; }
P operator*(const P& p, const P& q) { return P(p) *= q; }
P operator*(const P& p, const D& r) { return P(p) *= r; }
P operator/(const P& p, const D& r) { return P(p) /= r; }
P operator*(const D& r, const P& p) { return P(p) *= r; }
P operator/(const D& r, const P& p) { return P(p) /= r; }
D crs(const P& a, const P& b){ return a.x*b.y - a.y*b.x; }
D dot(const P& a, const P& b){ return a.x*b.x + a.y*b.y; }
int signum(D x) {return x > EPS ? +1 : x < -EPS ? -1 : 0;}
// 辞書順ソート
bool operator<(const P& a, const P& b) {
  if (a.x != b.x) return a.x < b.x;
  return a.y < b.y;
}


int main2() {
  int X = nextLong();

  P p(0, 0);
  P v(1, 0);

  int ans = 0;
  for (;;) {
    // cout << p << endl;
    p = p + v;
    v = v * P::polar(1.0, X/180.0 * PI);
    ans++;
    if (p.norm() < EPS)
      break;
  }
  cout << ans << endl;
  return 0;
}

int main() {

  // for (int i = 1; i <= 179; i++)
    // main2(i);
  // return 0;

#ifdef LOCAL
  for (;!cin.eof();cin>>ws)
#endif
    main2();
  return 0;
}