#include <iostream>
using namespace std;

int main(void){
  int N, M, K;
  cin >> N >> M >> K;
  bool ans = false;
  for(int n=0;n<=N;n++){
    for(int m=0;m<=M;m++){
      if((N-n)*m+(M-m)*n == K){
        ans = true;
      }
    }
  }
  cout << (ans ? "Yes" : "No") << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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 <iostream>
using namespace std;

int main(void){
  int N, M, K;
  cin >> N >> M >> K;
  bool ans = false;
  for(int n=0;n<=N;n++){
    for(int m=0;m<=M;m++){
      if((N-n)*m+(M-m)*n == K){
        ans = true;
      }
    }
  }
  cout << (ans ? "Yes" : "No") << endl;
  return 0;
}

```