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

int main(){
  int64_t N,M,K;
  cin>>N>>M>>K;
  for(int64_t i=0;i<=N;i++)
    for(int64_t j=0;j<=M;j++)
      if(i*(M-j)+j*(N-i)==K){
        cout<<"Yes"<<endl;
        return 0;
      }
  cout<<"No"<<endl;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

We have a grid with N rows and M columns of squares. Initially, all the squares are white.

There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted.

Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.

Constraints

* 1 \leq N,M \leq 1000
* 0 \leq K \leq NM

Input

Input is given from Standard Input in the following format:


N M K


Output

If Takahashi can have exactly K black squares in the grid, print `Yes`; otherwise, print `No`.

Examples

Input

2 2 2


Output

Yes


Input

2 2 1


Output

No


Input

3 5 8


Output

Yes


Input

7 9 20


Output

No

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int main(){
  int64_t N,M,K;
  cin>>N>>M>>K;
  for(int64_t i=0;i<=N;i++)
    for(int64_t j=0;j<=M;j++)
      if(i*(M-j)+j*(N-i)==K){
        cout<<"Yes"<<endl;
        return 0;
      }
  cout<<"No"<<endl;
}
```