Problem:
Score : 200 points

Problem Statement

La Confiserie d'ABC sells cakes at 4 dollars each and doughnuts at 7 dollars each.
Determine if there is a way to buy some of them for exactly N dollars. You can buy two or more doughnuts and two or more cakes, and you can also choose to buy zero doughnuts or zero cakes.

Constraints

N is an integer between 1 and 100, inclusive.

Input

Input is given from Standard Input in the following format:

N

Output

If there is a way to buy some cakes and some doughnuts for exactly N dollars, print Yes; otherwise, print No.

Sample Input 1

11

Sample Output 1

Yes

If you buy one cake and one doughnut, the total will be 4 + 7 = 11 dollars.

Sample Input 2

40

Sample Output 2

Yes

If you buy ten cakes, the total will be 4 \times 10 = 40 dollars.

Sample Input 3

3

Sample Output 3

No

The prices of cakes (4 dollars) and doughnuts (7 dollars) are both higher than 3 dollars, so there is no such way.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
#define ALL(a) a.begin(), a.end()
#define fi first
#define se second
#define endl "\n"
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

//const ll MOD = 998244353;
const ll MOD = (1e+9) + 7;
const ll INF = 2e+9 + 10;
typedef pair<ll, ll> P;


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N; cin >> N;
  rep(i, 100) {
    rep(j, 100) {
      if(4*i+7*j == N){
        cout << "Yes" << endl; return 0;
      }
    }
  }
  cout << "No" << endl;
  return 0;
}