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;
#define Rep(i,n) for(int i=0;i<(int)(n);i++)
#define For(i,n1,n2) for(int i=(int)(n1);i<(int)(n2);i++)
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define RREP(i,n) for(ll i=((ll)(n)-1);i>=0;i--)
#define FOR(i,n1,n2) for(ll i=(ll)(n1);i<(ll)(n2);i++)
#define put(a) cout<<a<<"\n"
#define all(a)  (a).begin(),(a).end()
#define SORT(a) sort((a).begin(),(a).end())
#define oorret 0
#define oor(x) [&](){try{x;} catch(const out_of_range& oor){return oorret;} return x;}()
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
template<typename T1, typename T2> inline bool chmin(T1& a, T2 b) { if (a > b) { a = b; return 1; }return 0; }
template<typename T1, typename T2> inline bool chmax(T1& a, T2 b) { if (a < b) { a = b; return 1; }return 0; }

template<typename T, typename U>
inline T pow(T x, U exp) {
	if (exp <= 0) {
		return 1;
	}
	if (exp % 2 == 0) {
		T d = pow(x, exp / 2);
		return d * d;
	}
	else {
		return (x * pow(x, exp - 1));
	}
}

template<typename T>
inline T fact(T n, vector<T>& table) {
	if (n >= (int)table.size()) {
		ll s = table.size();
		for (T i = s; i < n + 1; ++i) {
			table.push_back(table.back() * i);
		}
	}
	if (n < 0) return 1;
	else return table[n.a];
}

template<typename T>
inline T comb(T n, T m, vector<T>& table) {//nCm
	if (n - m < m)return comb(n, n - m, table);
	else return fact(n, table) / fact(m, table) / fact(n - m, table);
}

class ModInt {
public:
	ll a;
	const int mod = 1e9 + 7;
	ModInt(ll _val = 0) {
		if (_val >= mod) { _val %= mod; }
		a = _val;
	}
	ModInt operator=(const ModInt n) { a = n.a; return a; }
	ModInt operator+(const ModInt n) { if ((a + n.a) >= mod) { return a + n.a - mod; } else { return a + n.a; } }
	ModInt operator-(const ModInt n) { return a + (-n.a); }
	ModInt operator*(const ModInt n) { return a * n.a; }
	ModInt operator/(const ModInt n) { return (*this) * pow(n, mod - 2); }
	ModInt& operator+=(const ModInt n) { (*this) = (*this) + n; return *this; }
	ModInt& operator-=(const ModInt n) { (*this) = (*this) - n; return *this; }
	ModInt& operator*=(const ModInt n) { (*this) = (*this) * n; return *this; }
	ModInt& operator/=(const ModInt n) { (*this) = (*this) / n; return *this; }
	ModInt& operator++(int) { (*this) = (*this) + 1; return *this; }//前置インクリメントs(++a)のオーバーロード
	ModInt& operator++() { (*this) = (*this) + 1; return *this; }//後置インクリメント(a++)のオーバーロード
	ModInt& operator--(int) { (*this) = (*this) - 1; return *this; }//前置デクリメント(--a)のオーバーロード
	ModInt& operator--() { (*this) = (*this) - 1; return *this; }//後置デクリメント(a--)のオーバーロード
	ModInt operator-() { return mod - a; }//単項-演算子(-a)のオーバーロード
	ModInt inv() { ModInt temp(1); return temp / (*this); }//逆数を返す関数 return (*this)/(*this)/(*this);でもいい
	bool operator<(const ModInt n) { return a < n.a; }
	bool operator<=(const ModInt n) { return a <= n.a; }
	bool operator>(const ModInt n) { return a > n.a; }
	bool operator>=(const ModInt n) { return a >= n.a; }
	bool operator==(const ModInt n) { return a == n.a; }
	//下の関係演算子はpow関数で要請される
	bool operator<(const int n) { return a < n; }
	bool operator<=(const int n) { return a <= n; }
	bool operator>(const int n) { return a > n; }
	bool operator>=(const int n) { return a >= n; }
	bool operator==(const int n) { return a == n; }
	ModInt operator%(const int n) { return a % n; }
};
ostream& operator <<(ostream& o, const ModInt& t) {
	o << t.a;
	return o;
}
istream& operator >>(istream& i, ModInt& t) {
	i >> t.a;
	return i;
}


int main() {
	ll n,m,k;
	cin >> n >> m >> k;
	ModInt mi;
	REP(i, m) {
		mi += (m - 1 - i) * (m - i) / 2 + (i) * (i + 1) / 2;
	}
	mi *= n*n;
	ModInt mi2;
	REP(i, n) {
		mi2 += (n - 1 - i) * (n - i) / 2 + (i) * (i + 1) / 2;
	}
	mi2 *= m*m;
	mi /= 2;
	mi2 /= 2;
	mi += mi2;
	vector<ModInt> table(1, 1);
	ModInt temp(k - 2);
	ModInt temp2(n*m-2);
	mi *= comb(temp2, temp, table);
	put(mi);
	return 0;
}