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> //色々大嘘
//二つ目違うし比較も違う
using namespace std;
//関数
const long INF=1000000000000000;
const long MOD=1000000007;
bool check=false;

long modinv(long a) {
    long  b = MOD, u = 1, v = 0;
    while (b) {
        long  t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= MOD;
    if (u < 0) u += MOD;
    return u;
}

template< typename T >
T binomial(T N, T K) {
  if(K < 0 || N < K) return 0;
  T ret = 1;
  for(T i = 1; i <= K; ++i) {
    ret *= N--;
    ret%=MOD;
    ret *=modinv(i);
    ret%=MOD;
  }
  return ret%MOD;
}

int main(void){
    //cout<<binomial((long)10,(long)3)<<endl;
    long N,M,K;
    cin>>N>>M>>K;
    long ans=0,kasami=binomial(N*M-2,K-2);
    for(long d=1;d<N;d++){
        long pat=(N-d)*M*M;
        pat%=MOD;
        
        pat*=kasami;
        pat%=MOD;
        
        ans+=d*pat;
        ans%=MOD;
    }
    for(long d=1;d<M;d++){
        long pat=(M-d)*N*N;
        pat%=MOD;
        
        pat*=kasami;
        pat%=MOD;
        
        ans+=d*pat;
        ans%=MOD;
    }
    
    cout<<ans<<endl;
}