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 (1.15.1) (judge verdict: Wrong Answer)

#![allow(unused_mut)]
#![allow(non_snake_case)]
#![allow(unused_imports)]
use std::collections::HashSet;
use std::collections::HashMap;
use std::collections::BTreeSet;
use std::collections::VecDeque;
use std::cmp::{max, min};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}


// 参考
// https://atcoder.jp/contests/abc127/submissions/5626630

const MOD: usize = 1e9 as usize + 7;
const MAX: usize = 1e5 as usize * 2 + 1;

pub struct ModDivision {
    fact: Vec<usize>,
    inv: Vec<usize>,
    finv: Vec<usize>,
}

impl ModDivision {
    pub fn new(max_value: usize) -> Self {
        let mut fact = vec![0; max_value + 1];
        let mut inv = vec![0; max_value + 1];
        let mut finv = vec![0; max_value + 1];
        fact[0] = 1;
        fact[1] = 1;
        finv[0] = 1;
        finv[1] = 1;
        inv[1] = 1;
        for i in 2..max_value + 1 {
            fact[i] = fact[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
        ModDivision {
            fact: fact,
            inv: inv,
            finv: finv,
        }
    }

    pub fn combination(&self, n: usize, k: usize) -> usize {
        assert!(n >= k);
        self.fact[n] * self.finv[n - k] % MOD * self.finv[k] % MOD
    }
}

fn main() {
    input! {
        n: usize,
        m: usize,
        k: usize
    }
    let mut ans = 0;
    let div = ModDivision::new(MAX);
    let pair = div.combination(n * m - 2, k - 2);
    for d in 1..m {
        ans += n * n * (m - d) * d % MOD * pair % MOD;
        ans %= MOD;
    }
    for d in 1..n {
        ans += m * m * (n - d) * d % MOD * pair * MOD;
        ans %= MOD;
    }
    println!("{}", ans);
}