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: Accepted)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

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

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

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

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

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

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

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

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

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

struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut parent = vec![0; n];
        for i in 0..n {
            parent[i] = i;
        }
        return UnionFind {
            parent: parent,
            rank: vec![0; n],
            size: vec![1; n],
        };
    }

    fn find(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            let parent = self.parent[x];
            let pr = self.find(parent);
            self.parent[x] = pr;
            pr
        }
    }

    fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    fn unite(&mut self, a: usize, b: usize) {
        let a_root = self.find(a);
        let b_root = self.find(b);
        if self.rank[a_root] > self.rank[b_root] {
            self.parent[b_root] = a_root;
            self.size[a_root] += self.size[b_root];
        } else {
            self.parent[a_root] = b_root;
            self.size[b_root] += self.size[a_root];
            if self.rank[a_root] == self.rank[b_root] {
                self.rank[b_root] += 1;
            }
        }
    }

    fn get_size(&mut self, x: usize) -> usize {
        let root = self.find(x);
        self.size[root]
    }
}

const p: usize = 1000000007;

// fact(n) = n! mod p
fn fact(n: usize) -> usize {
    let mut acc = 1;
    for i in 1..n + 1 {
        acc = acc * i % p;
    }
    acc
}

fn mod_pow(b: usize, mut e: usize) -> usize {
    let mut base = b;
    let mut acc = 1;

    while e > 1 {
        if e % 2 == 1 {
            acc = acc * base % p;
        }
        e /= 2;
        base = base * base % p;
    }

    if e == 1 {
        acc = acc * base % p;
    }

    acc
}

fn comb(n: usize, r: usize) -> usize {
    // nCr = n! / (r! (n-r)!) = n! (r!)^(p-2) ((n-r)!)^(p-2)
    fact(n) * mod_pow(fact(r), p - 2) % p * mod_pow(fact(n - r), p - 2) % p
}

fn main() {
    input! {
        n: usize,
        m: usize,
        k: usize,
    }

    let mut ans = 0;

    for i in 1..n {
        ans += i * (n - i) * m * m;
        ans %= p;
    }
    for i in 1..m {
        ans += i * (m - i) * n * n;
        ans %= p;
    }

    ans = ans * comb(n * m - 2, k - 2) % p;

    println!("{}", ans);
}