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 ll long long 
#define pb push_back
#define mod 1000000007
#define all(v) v.begin(),v.end()
#define MP make_pair
#define F first
#define S second
#define inf 10000000000000
#define vll vector<ll>
using namespace std;
const int N=200005;
ll fact[N];
ll expo(ll x,ll y,ll m=mod){
    ll res=1LL;
    while(y){
        if(y&1) res=(res*x)%m;
        x=(x*x)%m;
        y/=2;
    }
    return res;
}
ll ncr(ll n,ll r)
{
	ll res=fact[n];
	res=(res*expo(fact[r],mod-2))%mod;
	res=(res*expo(fact[n-r],mod-2))%mod;
	return res;
}
void solve()
{
	ll n,m,k,i,j,p,q,ans=0;
	cin>>n>>m>>k;
	fact[0]=1;
	for(i=1;i<=n*m;i++)
		fact[i]=(fact[i-1]*i)%mod;
	for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
			{
				ans=(ans+ncr(n*m-2,k-2)*(((i*i-(i*(i+1))/2)*m)%mod))%mod;
				ans=(ans+ncr(n*m-2,k-2)*(((j*j-(j*(j+1))/2)*n)%mod))%mod;

			}		
					
		
	//ans=(ans*expo(2,mod-2))%mod;
	cout<<ans;

}
 
int main() {
    
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	
    ll t=1;
    //cin>>t;
    while(t--)
        solve();
 
    
    return 0;
}