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;
const int INF = 2147483647;
const double PI = 3.141592653589793;
map<int, int> mapa;
int n, m, x, k;
int oblicz(int a, int b) {
  a = max(a, 0);
  b = max(b, 0);
  int k1, k2;
  if (b % 2 == 1)
    k1 = b / 2 + 1;
  else
    k1 = b / 2;
  k2 = b / 2;
  return k1 * (a / 2 + a % 2) + k2 * (a / 2);
}
int main() {
  scanf("%d %d %d", &n, &m, &x);
  k = 0;
  while (n > 0 && m > 0) {
    k++;
    mapa[k] = oblicz(n, m) - oblicz(n - 2, m - 2);
    n -= 2;
    m -= 2;
  }
  printf("%d\n", mapa[x]);
  return 0;
}
