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>
using namespace std;
typedef long long ll;
const ll MAX=200005;
const ll MOD=1000000007;
ll fac[MAX],finv[MAX],inv[MAX];
ll ans;
void COMinit(){
	fac[0]=fac[1]=1;
	finv[0]=finv[1]=1;
	inv[1]=1;
	for(int i=2;i<MAX;i++){
		fac[i]=fac[i-1]*i%MOD;
		inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
		finv[i]=finv[i-1]*inv[i]%MOD;
	}
}
ll COM(ll i,ll j){
	if(i<j) return 0;
	if(i<0||j<0) return 0;
	return fac[i]*(finv[j]*finv[i-j]%MOD)%MOD;
}
int main(void){
	COMinit();
    ll n,m,k; cin>>n>>m>>k;
    //cout<<"AA"<<endl;
    for(int i=1;i<=n-1;i++){
		if(n*m-2>=MOD){
		}else{
			ans=(ans+i*m*m*(n-i)*COM(n*m-2,k-2))%MOD;
		}
		//cout<<ans<<endl;
	}
	for(int i=1;i<=m-1;i++){
		if(n*m-2>=MOD){
		}else{
			ans=(ans+i*n*n*(m-i)*COM(n*m-2,k-2))%MOD;
		}
		//cout<<ans<<endl;
	}
	cout<<ans<<endl;
}