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: Time Limit Exceeded)


"""
xi,xjのペアにおいて、このペアは何通り出てくるか
sum(全通り)sum(i)sum(j){|xi-xj|+|yi-yj|}
和なので、どんな順序でもいい

sum(i)sum(j)sum(全通り){|xi-xj|}
sum(i)sum(j)sum(全通り){|yi-yj|}

あるi,jで固定して、sum(全通り){|xi-xj|}を求めるヤツ
固定したi,jが現れる回数は、

pat = NM-2_C_K-2 こいつはどこでも一緒

pat*sum(i)sum(j){|xi-xj|}を計算する
すべてのi,j(i>j)ペアにおいて、xi,xjの和は
sum(i=1,N){i*(N-i)}
"""
from math import factorial

N,M,K = map(int,input().split())
mod = 10**9+7
fa = 1
num = N*M - 2
# 固定したときの通り数
for i in range(num,1,-1):
    if num-i == K-2:
        break
    fa *= i
pat = fa//factorial(K-2)
# xi-xj
n_x = M**2*((N-1)*N*(N+1))//6
n_y = N**2*((M-1)*M*(M+1))//6
print(pat*(n_x+n_y)%mod)