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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {
  public static long MOD = (long)Math.pow(10, 9) + 7;
  public static long[] kai;
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    long n = sc.nextLong();
    long m = sc.nextLong();
    long k = sc.nextLong();
    long ans = 0;
    
    kai = new long[200001];
    kai[0] = 1;
    for(long i = 1; i < 200001; i++) {
      kai[(int)i] = (kai[(int)i - 1] * i) % MOD;
    }
    
    for(long i = 1; i < n; i++) {
      long kosuu = ((n - i) * m) % MOD;
      kosuu = (kosuu * m) % MOD;
      kosuu = (kosuu * i) % MOD;
      kosuu = (kosuu * com((n * m - 2), k - 2)) % MOD;
      ans = (ans + kosuu) % MOD;
    }

    for(long i = 1; i < m; i++) {
      long kosuu = ((m - i) * n) % MOD;
      kosuu = (kosuu * n) % MOD;
      kosuu = (kosuu * i) % MOD;
      kosuu = (kosuu * com((n * m - 2), k - 2)) % MOD;
      ans = (ans + kosuu) % MOD;
    }

    System.out.println(ans);
  }

  public static long func(long a, long x) {
    if(x == 0) {
      return 1;
    } else if(x == 1) {
      return a % MOD;
    } else {
      if(x % 2 == 0) {
        return (func(a, x / 2) * func(a, x / 2)) % MOD;
      } else {
        return (a * func(a, x - 1)) % MOD;
      }
    }
  }

  public static long com(long s, long t) {
    if(s >= t) {
      long t1 = func(kai[(int)t], MOD - 2);
      long t2 = func(kai[(int)(s - t)], MOD - 2);
      long t3 = (kai[(int)s] * t1) % MOD;
      return (t3 * t2) % MOD;
    } else {
      return 0;
    }
  }
}