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: Runtime Error)

#include "bits/stdc++.h"
using namespace std;
#define rep(i,a,n) for(ll i=a;i<n;i++)
#define ALL(s) s.begin(),s.end()
#define P pair<ll,ll>
#define vi vector<int>
#define vl vector<ll>
#define vvi vector<vector<int>>
#define vvl vector<vector<ll>>
#define print(n) cout<<n<<endl
const int M=100010;
const int MOD=1000000007;
const int inf=1000000007;
const long long INF=1000000000000000007;
using ll=long long;


int main(){
	ll n,m,k,ans=0;
	cin>>n>>m>>k;
	ll dp[2000][2000];
	dp[0][0]=3*k*k+5*k-4;
	rep(i,1,n)rep(j,1,m){
		dp[i][j]=dp[0][0]+6*(i+j);
		if(dp[i][j]>inf)dp[i][j]=dp[i][j]%inf;
	}
	cout<<dp[k][k]<<endl;
}