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

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,M,K,ok=0; cin >> N >> M >> K;
    for (int i=0;i<=N;++i)
        for (int j=0;j<=M;++j)
            if (i*j+(N-i)*(M-j)==K) ok=1;
    cout << (ok?"Yes":"No") << '\n';
}
### 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(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N,M,K,ok=0; cin >> N >> M >> K;
    for (int i=0;i<=N;++i)
        for (int j=0;j<=M;++j)
            if (i*j+(N-i)*(M-j)==K) ok=1;
    cout << (ok?"Yes":"No") << '\n';
}
```