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: Wrong Answer)

#include "bits/stdc++.h"

using namespace std;
using ll = long long;
const double pi = acos(-1);
#define FOR(i,a,b) for (ll i=(a),__last_##i=(b);i<__last_##i;i++)
#define RFOR(i,a,b) for (ll i=(b)-1,__last_##i=(a);i>=__last_##i;i--)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define __GET_MACRO3(_1, _2, _3, NAME, ...) NAME
#define rep(...) __GET_MACRO3(__VA_ARGS__, FOR, REP)(__VA_ARGS__)
#define rrep(...) __GET_MACRO3(__VA_ARGS__, RFOR, RREP)(__VA_ARGS__)
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
	REP(i, v.size()) { if (i)os << " "; os << v[i]; }return os;
}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v) {
	REP(i, v.size()) { if (i)os << endl; os << v[i]; }return os;
}

const ll INF = LLONG_MAX;
ll MOD = 1000000007;
ll _MOD = 1000000009;
double EPS = 1e-10;

#define int long long

inline void my_io() {

	std::ios::sync_with_stdio(false);

	std::cin.tie(0);

	cout << fixed << setprecision(16);

	//cout << setprecision(10) << scientific << ans << endl;
}

signed main() {
	ll n;

	cin >> n;

	REP(i, 26) {
		ll x = n - i * 4;
		if (x % 7 == 0) {
			cout << "Yes" << endl;
			return 0;
		}
	}

	cout << "No" << endl;
}