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: Time Limit Exceeded)

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
#include<deque>
#include<numeric>
#include<queue>
#include<stack>
#include<cstring>
#include<limits>
#include<functional>
#include<unordered_set>
#include<iomanip>
#include<cassert>
#include<regex>
#include<bitset>
#include<complex>
#include<chrono>
#include<random>
#define rep(i,a) for(int i=(int)0;i<(int)a;++i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(),x.end()
using ll=long long;
constexpr ll mod = 1e9 + 7;
constexpr ll INF = 1LL << 60;

ll gcd(ll n, ll m) {
	ll tmp;
	while (m!=0) {
		tmp = n % m;
		n = m;
		m = tmp;
	}
	return n;
}

ll lcm(ll n, ll m) {
	return abs(n * m) / gcd(n, m);//gl=xy
}
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
using namespace std;

template< int mod >
struct ModInt {
  int x;

  ModInt() : x(0) {}

  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

  ModInt &operator+=(const ModInt &p) {
    if((x += p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator-=(const ModInt &p) {
    if((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator*=(const ModInt &p) {
    x = (int) (1LL * x * p.x % mod);
    return *this;
  }

  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }

  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

  bool operator==(const ModInt &p) const { return x == p.x; }

  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) {
    return os << p.x;
  }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt< mod >(t);
    return (is);
  }

  static int get_mod() { return mod; }
};

using modint = ModInt< mod >;

template< typename T >
struct Combination {
  vector< T > _fact, _rfact, _inv;

  Combination(int sz) : _fact(sz + 1), _rfact(sz + 1), _inv(sz + 1) {
    _fact[0] = _rfact[sz] = _inv[0] = 1;
    for(int i = 1; i <= sz; i++) _fact[i] = _fact[i - 1] * i;
    _rfact[sz] /= _fact[sz];
    for(int i = sz - 1; i >= 0; i--) _rfact[i] = _rfact[i + 1] * (i + 1);
    for(int i = 1; i <= sz; i++) _inv[i] = _rfact[i] * _fact[i - 1];
  }

  inline T fact(int k) const { return _fact[k]; }

  inline T rfact(int k) const { return _rfact[k]; }

  inline T inv(int k) const { return _inv[k]; }

  T P(int n, int r) const {
    if(r < 0 || n < r) return 0;
    return fact(n) * rfact(n - r);
  }

  T C(int p, int q) const {
    if(q < 0 || p < q) return 0;
    return fact(p) * rfact(q) * rfact(p - q);
  }

  T H(int n, int r) const {
    if(n < 0 || r < 0) return (0);
    return r == 0 ? 1 : C(n + r - 1, r);
  }
};

using P=pair<ll,ll>;
void solve(){
    ll n,m,k;
    cin>>n>>m>>k;
    modint ans=0;
    for(ll i=1;i<=m-1;++i){
        for(ll j=i+1;j<=m;++j){
            ans+=n*n*(j-i);
        }
    }
    for(ll i=1;i<=n-1;++i){
        for(ll j=i+1;j<=n;++j){
            ans+=m*m*(j-i);
        }
    }
    Combination<modint>c(3e5);
    ans*=c.C(n*m-2,k-2);
    cout<<ans;
}

int main(){
	ios::sync_with_stdio(false);
  cin.tie(0);
	cout<<fixed<<setprecision(15);
  solve();
	return 0;
}