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 fi first
#define se second
using namespace std;
using ll = long long;

ll constexpr N = 200009;
ll f[N];
ll constexpr MOD = 1000000007;

ll mp(ll a, ll b) {
    if(b == 0)
        return 1;
    ll h = mp(a, b / 2);
    if(b & 1)
        return (((h * h) % MOD) * a) % MOD;
    return (h * h) % MOD;
}

int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    ll n, m, k;
    cin >> n >> m >> k;
    vector<vector<ll>> dp(n + 10, vector<ll>(m + 10)),
                       dp2(n + 10, vector<ll>(m + 10));
    
    ll sum = 0;
    for(ll i = 1; i <= n; ++i) {
        for(ll j = 1; j <= m; ++j) {
            dp2[i][j] = j * (j - 1) / 2 + dp2[i - 1][j] + (i - 1) * j;

            dp[i][j] += dp2[i - 1][j] + (i - 1) * j;
            dp[i][j] %= MOD;
            dp[i][j] += dp2[i - 1][m - j] + 2 * (i - 1) * (m - j);
            dp[i][j] %= MOD;            
            dp[i][j] += j * (j - 1) / 2;
            dp[i][j] %= MOD;
            sum += dp[i][j];
            sum %= MOD;
        }
    }
    f[0] = 1;
    for(ll i = 1; i < N; ++i) {
        f[i] = (f[i - 1] * i) % MOD;
    }

    ll x = m * n - 2;
    ll y = k - 2;
    ll ans = sum * f[x];
    ans %= MOD;
    ans *= mp(f[y], MOD - 2);

    ans %= MOD;
    ans *= mp(f[x - y], MOD - 2);
    ans %= MOD;


    cout << ans << endl;
}