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())
import sys
mod = pow(10, 9) + 7
sys.setrecursionlimit(pow(10, 8))

def power(x, y):
    if   y == 0: return 1
    elif y == 1	 : return x % mod
    elif y % 2 == 0 : return power(x, y//2)**2 % mod
    else: return power(x, (y-1)//2)**2 * x % mod
    
def mul(a, b):
    return ((a % mod) * (b % mod)) % mod

def div(a, b):
    return mul(a, power(b, mod-2))
def div2(a, b):
    return mul(a, modinv(b))

def modinv(a):
    b, u, v = mod, 1, 0
    while b:
        t = a//b
        a, u = a-t*b, u-t*v
        a, b, u, v = b, a, v, u
    u %= mod
    return u

def cmb(n, r):
    if ( r<0 or r>n ):
        return 0
    r = min(r, n-r)
    return g1[n] * g2[r] * g2[n-r] % mod

NNN = (10**6)
g1 = [1, 1]
g2 = [1, 1]
inverse = [0, 1]

for i in range( 2, NNN + 1 ):
    g1.append( ( g1[-1] * i ) % mod )
    inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
    g2.append( (g2[-1] * inverse[-1]) % mod )

def f(n, m):
    x = m*m%mod
    y = (n*n*(n-1)//2 - n*(n-1)*(2*n-1)//6) % mod
    return x*y%mod

print(mul(f(N, M) + f(M, N), cmb(N*M-2, K-2)))