Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

2 2 2

Sample Output 1

8

There are six possible arrangements of the pieces, as follows:

((1,1),(1,2)), with the cost |1-1|+|1-2| = 1

((1,1),(2,1)), with the cost |1-2|+|1-1| = 1

((1,1),(2,2)), with the cost |1-2|+|1-2| = 2

((1,2),(2,1)), with the cost |1-2|+|2-1| = 2

((1,2),(2,2)), with the cost |1-2|+|2-2| = 1

((2,1),(2,2)), with the cost |2-2|+|1-2| = 1

The sum of these costs is 8.

Sample Input 2

4 5 4

Sample Output 2

87210

Sample Input 3

100 100 5000

Sample Output 3

817260251

Be sure to print the sum modulo 10^9+7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MOD = (int) 1e9 + 7;

int add(int a, int b) {
        a += b;
        if (a >= MOD) {
                return a - MOD;
        }
        if (a < 0) {
                return a + MOD;
        }
        return a;
}

int mul(int a, int b) {
        return a * (ll) b % MOD;
}

int expow(int a, int b) {
        int r = 1;
        while (b) {
                if (b & 1) {
                        r = mul(r, a);
                }
                a = mul(a, a);
                b >>= 1;
        }
        return r;
}

int get_factorial(int i) {
        if (i == 0) {
                return 1;
        } else {
                return mul(get_factorial(i - 1), i);
        }
}

int inv(int x) {
        return expow(x, MOD - 2);
}

int comb(int n, int k) {
        int x = get_factorial(n), y = inv(mul(get_factorial(k), get_factorial(n - k)));
        return mul(x, y);
}

int sqr(int x) {
        return mul(x, x);
}

int n, m, k;

int main() {
        cin >> n >> m >> k;
        int a = comb(n * m - 2, k - 2), b = 0;

        for (int dr = 1; dr < n; dr++) {
                int pick = n - dr;
                b = add(b, mul(dr, mul(pick, sqr(m))));
        }
        for (int dc = 1; dc < m; dc++) {
                int pick = m - dc;
                b = add(b, mul(dc, mul(pick, sqr(n))));
        }

        cout << mul(a, b) << "\n";

        return 0;
}