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;
int n, m, x;
vector<int> v;
int main() {
  cin >> n >> m >> x;
  if (x > n / 2 + 1)
    cout << 0;
  else {
    for (int i = 2; i <= x; i++) n -= 2, m -= 2;
    if (n == 1 && m == 1) {
      cout << 1;
      return 0;
    }
    if (n <= 0 || m <= 0) {
      cout << 0;
      return 0;
    }
    int tx1 = n / 2 + n % 2, tx2;
    tx2 = tx1;
    if (m % 2 == 0 && n % 2 == 1) tx2--;
    int ty1, ty2;
    ty1 = m / 2 + m % 2 - 1;
    if (m % 2 == 1) ty1--;
    if (n % 2 == 1)
      ty2 = m / 2 + m % 2;
    else
      ty2 = m / 2 + m % 2 - 1;
    if (n % 2 == 1) ty2--;
    if (m % 2 == 1 && n % 2 == 1) ty2--;
    cout << tx1 + tx2 + ty1 + ty2;
  }
}
