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)

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

const int N = 2e5 + 5;
int n,m,k;

using ll = long long;
const int mod = 1e9 + 7;
int fac[N], finv[N];

ll quick_pow(ll a, ll b) {
    ll res = 1;
    a %= mod;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

void init_fac_mod(int to){
    //only when mod is prime
    fac[0] = finv[0] = 1;
    for(int i = 1; i <= to; ++i)
        fac[i]=(ll)fac[i-1]*i%mod;
    finv[to] = quick_pow(fac[to], mod-2);
    for(int i = to-1; i; --i)
        finv[i] = (ll)finv[i+1]*(i+1)%mod;
}

int comb(int n, int m){
    //n!/(m!(n-m)!)
    init_fac_mod(n);
    return (ll)fac[n] * finv[m] % mod * finv[n - m]% mod;
}



int main(){
#ifdef LOCAL
    freopen("./input.txt", "r", stdin);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    
    cin>>n>>m>>k;
    
    
    ll d = 0;
    for (int i = 1; i < n; i++){
        d += (ll)i*(n-i)%mod*m%mod*m%mod;
        d%= mod;
    }
    
    for (int i = 1; i < m; i++){
        d += (ll)i*(m-i)%mod*n%mod*n%mod;
        d %= mod;
    }
    
    cout<<d*comb(n*m-2, k-2)%mod<<endl;
    
    
    
    
    
    return 0;
}