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.

Examples

Input

2 2 2


Output

8


Input

4 5 4


Output

87210


Input

100 100 5000


Output

817260251
#include <bits/stdc++.h>
#define ll long long int
#define MOD 1000000007
#define INF 1000000000000000000

using namespace std;
ll power(ll n, ll k){
  ll res = 1;
  while(k > 0){
    if (k & 1) 
      res = res * n % MOD;
    n = n * n % MOD;
    k >>= 1;
  }
  return res;
}

ll genfact(ll n, ll k){

  ll res = 1;
  k--;
  while (k >= 0){
    res = (res * (n - k)) % MOD;
    k--;
  }

  return res;
}

ll inv(ll n){
  return power(n, MOD - 2);
}

ll fact(ll n){
  return genfact(n, n - 1);
}

ll nck(ll n, ll k){
  if (k == 0)
    return 1;
  if (n <= 0)
    return 0;

  return ((genfact(n, k) % MOD) * inv(fact(k))) % MOD;
}

int main(void){
  ll n, m, k;
  cin >> n >> m >> k;
  
  ll ans = 0;

  ll t = nck(n*m-2,k-2);
  for (ll j = 1; j < m; j++){
    ans += j*(m-j) % MOD * n % MOD *n % MOD * t % MOD;
    ans = (ans % MOD);
  }
  for (ll i = 1; i < n; i++){
    ans += i*(n-i) % MOD * m % MOD *m % MOD * t % MOD;
    ans = (ans % MOD);
  }
  cout << ans << endl;

  return 0;
}
