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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    unused_imports,
    dead_code
)]
//use proconio::fastout;
use proconio::input;
use proconio::marker::*;
use std::cmp::*;
use std::collections::*;
//use std::collections::VecDeque;

const MOD: i64 = 1_000_000_007;
fn main() {
    input! {
       N:i64,M:i64,K:i64
    }
    // xの差で考える。
    // 1次元なら、xの差は1がM-1、2がM-2通り...。
    // 2次元なら、各差分について、N*N通り。
    let mut ans = 0i64;
    for i in 0..M {
        ans += i * (M - i) * (N * N);
    }
    for i in 0..N {
        ans += i * (N - i) * (M * M);
    }
    ans %= MOD;
    let mut fact = 1;
    for i in 1..=(K - 2) {
        fact = fact * i % MOD;
    }
    let ifactK_2 = modInv(fact);
    let mut fact = 1;
    for i in (N * M - K + 1)..=(N * M - 2) {
        fact = fact * i % MOD;
    }
    let factNM_2 = fact;
    let mut pat = factNM_2 * ifactK_2 % MOD;
    ans *= pat;
    ans = (ans + MOD) % MOD;
    println!("{}", ans);
}
fn modInv(a: i64) -> i64 {
    modPow(a, MOD - 2) % MOD
}
fn modPow(x: i64, a: i64) -> i64 {
    if a == 1 {
        x
    } else if a % 2 != 0 {
        x * modPow(x, a - 1) % MOD
    } else {
        modPow(x, a / 2).pow(2) % MOD
    }
}