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 Python (3.4.3) (judge verdict: Accepted)

def egcd(a, b):
    (x, lastx) = (0, 1)
    (y, lasty) = (1, 0)
    while b != 0:
        q = a // b
        (a, b) = (b, a % b)
        (x, lastx) = (lastx - q * x, x)
        (y, lasty) = (lasty - q * y, y)
    return (lastx, lasty, a)

# ax ≡ 1 (mod m)なるxを返す
def modinv(a):
    m=MOD
    (inv, q, gcd_val) = egcd(a, m)
    return inv % m

MOD=10**9+7
N,M,K=map(int,input().split())

b=1
for i in range(N*M-2,N*M-K,-1):
    b=(b*i)%MOD

for i in range(1,K-1):
    b=(b*modinv(i))%MOD

x=0
for i in range(1,N):
   x=x+i*(N-i)

x=(x*M*M)%MOD

y=0
for j in range(1,M):
    y=y+j*(M-j)
y=(y*N*N)%MOD
    
ans=b*(x+y)%MOD
print(ans)