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)

/* nothing is impossible */
#pragma comment(linker, "/stack:227420978")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h> 
 
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
using namespace std;
 
typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> pii;
typedef double ld;
typedef string st;
const double PI=3.14159265358979323846264338327950288419716939937510582097494459230;
const ll MOD = 1e9 + 7 ;
const ll INF=1e14;                                                                                                 
ll mpow(ll a,ll b,ll p=MOD){a=a%p;ll res=1;while(b>0){if(b&1)res=(res*a)%p;a=(a*a)%p;b=b>>1;}return res%p;}             

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
 // freopen("input.txt", "r", stdin);
  ll n;
  cin>>n;
  ll i,j;
  bool f=false;
  for(i=0;i<104;i++)
  {
    for(j=0;j<104;j++)
    {
      if(i*4+7*j==n)
      {
        f=true;
        break;
      }
    }
    if(f) break;
  }
  if(f)
  {
    //if(i==0||j==0||i>2&&j>2) cout<<"YES";
    //else 
      cout<<"Yes";
  }
  else cout<<"No";
return 0;
}