A chessboard n × m in size is given. During the zero minute we repaint all the black squares to the 0 color. During the i-th minute we repaint to the i color the initially black squares that have exactly four corner-adjacent squares painted i - 1 (all such squares are repainted simultaneously). This process continues ad infinitum. You have to figure out how many squares we repainted exactly x times.

The upper left square of the board has to be assumed to be always black. Two squares are called corner-adjacent, if they have exactly one common point.

Input

The first line contains integers n and m (1 ≤ n, m ≤ 5000). The second line contains integer x (1 ≤ x ≤ 109).

Output

Print how many squares will be painted exactly x times.

Examples

Input

3 3
1


Output

4


Input

3 3
2


Output

1


Input

1 1
1


Output

1
#include <bits/stdc++.h>
using namespace std;
bool arr[5005][5005];
int main() {
  int i, a, n, m, x, ret = 0;
  scanf("%d %d %d", &n, &m, &x);
  n -= 2 * (x - 1);
  m -= 2 * (x - 1);
  if (n <= 0 && m <= 0)
    printf("0\n");
  else {
    for (i = 0; i < n; i++) {
      if (i % 2 == 0)
        arr[i][0] = 1;
      else
        arr[i][0] = 0;
      for (a = 1; a < m; a++) arr[i][a] = !arr[i][a - 1];
    }
    for (i = 0; i < n; i++)
      for (a = 0; a < m; a++)
        if (i == 0 || i == n - 1 || a == 0 || a == m - 1) ret += arr[i][a];
    printf("%d\n", ret);
  }
  return 0;
}
