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<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const ll MOD=1e9+7;
ll n,m,k,fact[200005],inv[200005];
ll quick_pow(ll x,ll a)
{
	ll ans=1;
	while(a)
	{
		if(a&1)ans=ans*x%MOD;
		x=x*x%MOD;
		a>>=1;
	}
	return ans;
}
ll C(int n,int m)
{
	if(n<m)return 0;
	return fact[n]*inv[m]%MOD*inv[n-m]%MOD;
}
int main()
{
	scanf("%lld%lld%lld",&n,&m,&k);
	fact[0]=1;
	for(int i=1;i<=n*m;i++)fact[i]=fact[i-1]*i%MOD;
	inv[n*m]=quick_pow(fact[n*m],MOD-2);
	for(int i=n*m;i>=1;i--)inv[i-1]=inv[i]*i%MOD;
	ll ans=0;
	for(int i=1;i<n;i++)
		ans+=C(n*m-2,k-2)*i*m%MOD*(n-i)*m%MOD;
	swap(n,m);
	for(int i=1;i<n;i++)
		ans+=C(n*m-2,k-2)*i*m%MOD*(n-i)*m%MOD;
	printf("%lld\n",ans%MOD);
	return 0;
}