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
import java.util.Scanner;


public class B {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int x = sc.nextInt();
        for (int i = 1; i < x; i++) {
            n -= 2;
            m -= 2;
            if (n <= 0 || m <= 0) {
                System.out.println(0);
                return;
            }
        }
        int ans = calc(n, m)-calc(n-2, m-2);
        System.out.println(ans);
    }

    private static int calc(int n, int m) {
        if (n <= 0 || m <= 0)
            return 0;
        int ans = 0;
        if (n % 2==0)
            ans = n/2*m;
        else {
            if (m % 2==0)
                ans = m/2*n;
            else
                ans = (n/2+1)*(m/2+1)+(n/2)*(m/2);
        }
        return ans;
    }

}
