#include<iostream>
using namespace std;
int main()
{
int N;
int judge=0;
cin>>N;
for(int i=0;i<=100;i=i+4)
{
for(int j=0;j<=100;j=j+7)
{
if(N==i+j) judge++; 
}
}
if(judge>0) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}

### Prompt

Develop a solution in CPP to the problem described below:

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`.

Examples

Input

11


Output

Yes


Input

40


Output

Yes


Input

3


Output

No

### Response

```cpp
#include<iostream>
using namespace std;
int main()
{
int N;
int judge=0;
cin>>N;
for(int i=0;i<=100;i=i+4)
{
for(int j=0;j<=100;j=j+7)
{
if(N==i+j) judge++; 
}
}
if(judge>0) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}

```