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;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
typedef vector<ll> vll;
typedef vector<ld> vld;
typedef vector<pll> vpl;
typedef vector<vll> vvll;

#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define FI first
#define SE second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define REP1(i,n) for(int i=1;i<((int)n);i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define PB push_back
#define EB emplace_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define yes cout<<"Yes"<<endl
#define YES cout<<"YES"<<endl
#define no cout<<"No"<<endl
#define NO cout<<"NO"<<endl
#define Decimal fixed<<setprecision(20)
#define INF 1000000000
#define LLINF 1000000000000000000LL

const int inf = 1e9;
const ll linf = 1LL << 50;
const double eps = 1e-10;
const int MOD = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};


ll mpow(ll m, ll p){
  ll power = m;
  ll res = 1;
  while(p>0){
    if(p&1)
      res = res*power%MOD;
    power = (power*power)%MOD;
    p /= 2;
  }

  return res;
}

ll mod_inv(ll m){
  return mpow(m, MOD-2);
}

struct COM{
  vll fact, fact_inv;
  COM(ll n): fact(n+1,1), fact_inv(n+1,1){
    for(ll i=1; i<=n; i++)
      fact[i] = fact[i-1]*i%MOD;

    fact_inv[n] = mod_inv(fact[n]);
    for(ll i=n; i>=1; i--)
      fact_inv[i-1] = fact_inv[i]*i%MOD;
  }

  ll calc(ll n, ll k){
    if(k<0 || n<k)
      return 0;
    ll res = fact[n]*fact_inv[n-k]%MOD*fact_inv[k]%MOD;
    return res;
  }
};

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  ll n,m,k;
  cin>>n>>m>>k;

  COM com(n*m);
  ll ans2=0;
  for(ll d=1; d<=m-1; d++){
    ll cost=d*n*n*(m-d)%MOD;
    ans2=(ans2+com.calc(n*m-2,k-2)*cost%MOD)%MOD;
  }
  for(ll d=1; d<=n-1; d++){
    ll cost=d*m*m*(n-d)%MOD;
    ans2=(ans2+com.calc(n*m-2,k-2)*cost%MOD)%MOD;
  }

  cout<<ans2<<endl;
}