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>

#define M_PI       3.14159265358979323846   // pi

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<ll, ll> P;

#define rep(a,n) for(int a = 0;a < n;a++)
#define repi(a,b,n) for(int a = b;a < n;a++)

const ull mod = 1000000007;

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

ll inv_mod(ll a) { return mpow(a, mod - 2); }

class Factorial {
private:
	vector<ll> fac;
	vector<ll> ifac;
public:

	Factorial(ll N) {
		fac.push_back(1);
		for (int i = 0; i < N; i++) fac.push_back(fac[i] * (i + 1) % mod);
		ifac.resize(N + 1);
		ifac[N] = inv_mod(fac[N]);
		for (int i = 0; i < N; i++) ifac[N - 1 - i] = (ifac[N - i] * (N - i)) % mod;
	}

	ll fact(ll a) { return fac[a]; }
	ll ifact(ll a) { return ifac[a]; }

	ll cmb(ll a, ll b) {
		if (a == 0 && b == 0) return 1;
		if (a < b || a < 0 || b < 0) return 0;
		ll tmp = ifact(a - b) * ifact(b) % mod;
		return tmp * fac[a] % mod;
	}
	ll per(ll a, ll b) {
		if (a == 0 && b == 0) return 1;
		if (a < b || a < 0 || b < 0) return 0;
		return fac[a] * ifac[a - b] % mod;
	}
};

int main(void)
{
	ll n, m, k;
	cin >> n >> m >> k;
	Factorial f(n * m);
	ll comb = f.cmb(n * m - 2, k - 2);
	ll x = 0;
	rep(i, n) {
		ll dist = i;
		ll count = n - i;
		ll select = m * m;
		ll num = (dist * count * select) % mod;
		x += (num * comb) % mod;
		x %= mod;
	}
	swap(n, m);
	rep(i, n) {
		ll dist = i;
		ll count = n - i;
		ll select = m * m;
		ll num = (dist * count * select) % mod;
		x += (num * comb) % mod;
		x %= mod;
	}
	cout << x << endl;
	return 0;
}