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 PyPy3 (2.4.0) (judge verdict: Accepted)

n, m, k = map(int, input().split())

mod = 1000000007
def pow(x, n):
    ret = 1
    while n > 0:
        if (n & 1) == 1:
            ret = (ret * x) % mod
        x = (x * x) % mod
        n //= 2
    return ret

fac = [1]
inv = [1]
for i in range(1, n * m + 1):
    fac.append((fac[-1] * i) % mod)
    inv.append(pow(fac[i], mod - 2))

def cmb(n, k):
    return ((fac[n] * inv[k]) % mod * inv[n - k]) % mod

def doit(n, m, k):
    ret = 0
    for d in range(m):
        ret = (ret + d * (m - d) * n * n * cmb(n * m - 2, k - 2)) % mod
    return ret

print((doit(n, m, k) + doit(m, n, k)) % mod)