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 mm(x,k) memset (x, k, sizeof(x));
#define ll long long
#define ff(i,a,b) for(ll i = a ; i  < b ; ++i)
#define fd(i,a,b) for(ll i = a ; i <= b ; ++i)
using namespace std;
#define mo 1000000007
ll mod(ll a , ll b){
    ll t = 1;
    while(b){
        if(b&1) t = a * t % mo;
        b >>= 1; a = a * a % mo;
    } return t;
}
ll a[200005];
ll C(ll n , ll m){
    return (a[n] * mod(a[n-m],mo-2)) % mo * mod(a[m],mo-2) %mo;
}
int main(){
    a[0] = 1;
    ff(i,1,200005) a[i] = a[i-1] * i % mo;
    ll n , m , k;
    cin >> n >> m >> k;
    cout<< C(n*m,k)*C(k,2)%mo*(n+m)%mo*mod(3,mo-2)%mo;
}