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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define rep(i,l,r) for((i)=(l);(i)<(r);(i)++)
#define abs(a) ((a)>(0)?(a):-(a))
#define MOD 1000000007
#define ll long long
ll inv(ll a){
  ll t=1;
  while (a>1){
    t=t*(MOD-MOD/a)%MOD;
    a=MOD%a;
  }
  return t;
}
ll comb(ll n,ll k){
  if (k>n-k) k=n-k;
  ll nu=1,de=1,i;
  rep(i,1,k+1){
    nu=nu*(n+1-i)%MOD;
    de=de*i%MOD;
  }
  return nu*inv(de)%MOD;
}
int main(){
    long long n,m,k,ans=0,i;
    scanf("%lld%lld%lld",&n,&m,&k);
    rep(i,1,n) ans+=i*abs(n-i)*(m*m);
    ans%=MOD;
    rep(i,1,m) ans+=i*abs(m-i)*(n*n);
    ans%=MOD;
    ans*=comb(n*m-2,k-2);
    printf("%lld",ans%MOD);
}