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>
using namespace std;
typedef long long ll;
//#include <boost/multiprecision/cpp_int.hpp>
//typedef boost::multiprecision::cpp_int ll;
typedef long double dd;
#define i_7 (ll)(1E9+7)
//#define i_7 998244353
#define i_5 i_7-2
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
ll inf=(ll)1E17;

#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
///////////////////////////////////////
#define N 500004

ll kai[N];
ll kai2[N];
ll po(ll i,ll p){
    if(p==0)return 1;
    else if(p==1)return mod(i);
    else if(p%2==0)return po(mod(mod(i)*mod(i)),p/2);
    else  return mod(mod(i)*po(i,p-1));
}
void calc(){
    kai[0]=1;
    kai2[0]=1;
    rep(i,1,N-1){
        kai[i]=mod(kai[i-1]*i);
    }
    kai2[N-1]=po(kai[N-1],i_5);
    for(ll i=N-2;i>=0;i--){
        kai2[i]=mod(kai2[i+1]*(i+1));
    }
}
ll comb(ll n,ll k){
    if(n<k)return 0;
    else if(n==0)return 1;
    else return mod(mod(kai[n]*kai2[n-k])*kai2[k]);
}
ll bunbo(ll n){
    return po(n,i_5);
}


int main(){fastio
    calc();
    ll n,m,k;cin>>n>>m>>k;
    ll last=kai2[k-2];
    ll nowu=mod(n*m-2);
    rep(i,1,k-2){
        last=mod(last*nowu);
        nowu--;
        if(nowu<0)nowu=mod(nowu);
    }
    ll s[N];s[0]=0;rep(i,1,N-1)s[i]=s[i-1]+i;
    ll sum1=0,sum2=0;
    rep(i,1,m){
        sum1+=s[i-1]+s[m-i];
    }
    sum1=mod(mod(sum1*n)*n);
    rep(i,1,n){
        sum2+=s[i-1]+s[n-i];
    }
    sum2=mod(mod(sum2*m)*m);
    
    //cout<<sum1<<" "<<sum2<<endl;
    sum1=mod(sum1+sum2);
    last=mod(last*sum1);
    last=mod(last*bunbo(2));
    cout<<last<<endl;
    
    return 0;
}