We have a grid of squares with N rows and M columns. Let (i, j) denote the square at the i-th row from the top and j-th column from the left. We will choose K of the squares and put a piece on each of them.

If we place the K pieces on squares (x_1, y_1), (x_2, y_2), ..., and (x_K, y_K), the cost of this arrangement is computed as:

\sum_{i=1}^{K-1} \sum_{j=i+1}^K (|x_i - x_j| + |y_i - y_j|)

Find the sum of the costs of all possible arrangements of the pieces. Since this value can be tremendous, print it modulo 10^9+7.

We consider two arrangements of the pieces different if and only if there is a square that contains a piece in one of the arrangements but not in the other.

Constraints

* 2 \leq N \times M \leq 2 \times 10^5
* 2 \leq K \leq N \times M
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M K


Output

Print the sum of the costs of all possible arrangements of the pieces, modulo 10^9+7.

Examples

Input

2 2 2


Output

8


Input

4 5 4


Output

87210


Input

100 100 5000


Output

817260251
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int m = stdin.nextInt();
        int k = stdin.nextInt();
        
        long ans = 0;
        int t = c(n * m - 2, k - 2);
        for (long i = 0; i < n; i++) ans = (ans + ((n - i) * m * m * i) % 1000000007) % 1000000007;
        for (long i = 0; i < m; i++) ans = (ans + ((m - i) * n * n * i) % 1000000007) % 1000000007;
        System.out.println((ans * t) % 1000000007);
    }
    
    private static List<BigInteger> fct = new ArrayList<>();
    private static List<BigInteger> inv = new ArrayList<>();
    private static BigInteger mod = BigInteger.valueOf(1000000007);
    private static int c(int n, int r) {
        if (fct.isEmpty()) {
            fct.add(BigInteger.ONE);
            inv.add(fct.get(0).modInverse(mod));
        }

        for (int i = fct.size(); i <= n; i++) {
            fct.add(fct.get(i - 1).multiply(BigInteger.valueOf(i)).mod(mod));
            inv.add(fct.get(i).modInverse(mod));
        }

        return fct.get(n).multiply(inv.get(n - r).multiply(inv.get(r)).mod(mod)).mod(mod).intValue();
    }
}