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.io.*;
import java.util.*;

public class Repaintings {
    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        int x = Integer.parseInt(f.readLine());
        if (x > 5000) {
            System.out.println(0);
            return;
        }
        n -= 2*(x-1);
        m -= 2*(x-1);
        
        System.out.println(f(n,m)-f(n-2,m-2));
    }
    
    public static int f(int n, int m) {
        if (n <= 0 || m <= 0)
            return 0;
        return (n*m+1)/2;
    }
}