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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<int> Vi;
typedef vector<ll> Vl;
typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;


//===

const ll mod = static_cast<ll>(1e9 + 7);


ll mpow(ll a, ll n) {
	ll res = 1;
	while (n != 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

ll mfrac(ll a) {
	ll res = 1;
	for (ll i = 2; i <= a; ++i) res = res * i % mod;
	return res;
}


ll mcomb(ll n, ll k) {
	ll res;
	res = mfrac(n) * mpow(mfrac(k), mod - 2) % mod;
	res = res * mpow(mfrac(n - k), mod - 2) % mod;
	return res;
}


int main() {
	ll N, M, K;
	cin >> N >> M >> K;

	vector<Pll> cells;
	for (ll i = 0; i < N * M; ++i) {
		cells.emplace_back(i / M, i % M);
	}

	ll ans = 0;

	for (ll i = 0; i < N * M - 1; ++i) {
		for (ll j = i + 1; j < N * M; ++j) {
			ll dx = abs(cells[i].first - cells[j].first);
			ll dy = abs(cells[i].second - cells[j].second);
			ll cmb = mcomb(N * M - 2, K - 2);
			ans += dx * cmb % mod;
			ans += dy * cmb % mod;
			ans %= mod;
		}
	}

	cout << ans << endl;
}