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: Wrong Answer)

use std::cmp::*;
use std::io::*;

mod enumeration {
    pub struct Enumeration {
        fact: Vec<u64>,
        finv: Vec<u64>,
        modulo: u64,
    }
    impl Enumeration {
        pub fn new(n: usize, modulo: u64) -> Enumeration {
            let mut fact: Vec<u64> = vec![1; n];
            let mut inv: Vec<u64> = vec![1; n];
            let mut finv: Vec<u64> = vec![1; n];
            for i in 2..n {
                fact[i] = fact[i - 1] * (i as u64) % modulo;
                inv[i] = modulo - (modulo / (i as u64)) * inv[(modulo as usize % i)] % modulo;
                finv[i] = inv[i] * finv[i - 1] % modulo;
            }
            Enumeration {
                fact: fact,
                finv: finv,
                modulo: modulo,
            }
        }
        pub fn c(&self, n: usize, k: usize) -> u64 {
            if n < k {
                return 0;
            }
            self.fact[n] * self.finv[n - k] % self.modulo * self.finv[k] % self.modulo
        }
    }
}
fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: u64 = itr.next().unwrap().parse().unwrap();
    let m: u64 = itr.next().unwrap().parse().unwrap();
    let k: usize = itr.next().unwrap().parse().unwrap();

    const MOD: u64 = 1_000_000_007;
    let e = enumeration::Enumeration::new((n * m) as usize, MOD);

    // x
    // 2点を固定する。
    // それ以外の選び方はn*m - 2 C k - 2
    // 大きさがxとなる　-> n - x通り * m * m

    let mut ans = 0;
    for x in 1..n {
        ans += x * (n - x) % MOD * m % MOD * m % MOD;
    }
    for x in 1..m {
        ans += x * (m - x) % MOD * n % MOD * n % MOD;
    }
    ans = ans * e.c((n * m) as usize - 2, k - 2) % MOD;
    println!("{}", ans);
}