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 N = 4e5 + 10, Mod = 1e9 + 7;
inline int Power(int a, int b)
{
    int ret = 1;
    for (; b; b >>= 1, a = (ll)a * a % Mod)
        if (b & 1) ret = (ll)ret * a % Mod;
    return (ret);
}
int main()
{
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);

    ll cn = 1;
    for (int i = 1; i <= k; i++)
        cn = cn * i % Mod;
    cn = Power(cn, Mod - 2);
    cn = cn * k % Mod * (k - 1) % Mod;
    for (int i = n * m - k + 1; i <= n * m - 2; i++)
        cn = cn * i % Mod;
    ll tot = 0;
    for (int i = 1; i < n; i++)
        tot = (tot + (ll)(n - i) * m % Mod * m % Mod * i) % Mod;
    for (int i = 1; i < m; i++)
        tot = (tot + (ll)(m - i) * n % Mod * n % Mod * i) % Mod;
    tot = tot * cn % Mod;
    printf("%lld\n", tot);
    return 0;
}