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 int ll;
ll MOD = 1000000007;
ll INFL = 1ll << 60;
ll INF = 1 << 28;

const long long MAX = 510000;  // 最大を決めておく
// fac…階乗、inv…逆数、finv…逆数の累積積
long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理、先にやっておく
void nCr_init() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

// nCrを計算する (前処理は忘れないように)
long long nCr(int n, int r) {
  if (n < r) return 0;
  if (n < 0 || r < 0) return 0;
  return fac[n] * (finv[r] * finv[n - r] % MOD) % MOD;
}

// ====================================================================

int main() {
  nCr_init();

  ll h, w, k;
  cin >> h >> w >> k;

  ll ans = 0;
  for (ll dx = 1; dx < w; dx++) {  // その横幅を、全探索。
    ans = (dx * (w - dx) * h % MOD * h % MOD * nCr(h * w - 2, k - 2) % MOD + ans) % MOD;
  }
  for (ll dy = 1; dy < h; dy++) {  // その横幅を、全探索。
    ans = (dy * (h - dy) * w % MOD * w % MOD * nCr(h * w - 2, k - 2) % MOD + ans) % MOD;
  }

  cout << ans << endl;
}