#include <stdio.h>

int main () {
    int i,j,m = 0;
    int N,M,K;
    scanf("%d %d %d",&N,&M,&K);
    for (i=0;i<=N;i++){
        for (j=0;j<=M;j++){
            if (j*(N - i)+i*(M - j) == K){m=1;}
        }
    }
    if (m==1){printf("Yes\n");}
    else{printf("No\n");}
    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 <stdio.h>

int main () {
    int i,j,m = 0;
    int N,M,K;
    scanf("%d %d %d",&N,&M,&K);
    for (i=0;i<=N;i++){
        for (j=0;j<=M;j++){
            if (j*(N - i)+i*(M - j) == K){m=1;}
        }
    }
    if (m==1){printf("Yes\n");}
    else{printf("No\n");}
    return 0;
}
```