Problem:
Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.

Find the inversion number of B, modulo 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

Thus, the inversion number of B is 3.

Sample Input 2

3 5
1 1 1

Sample Output 2

0

A may contain multiple occurrences of the same number.

Sample Input 3

10 998244353
10 9 8 7 5 6 3 4 2 1

Sample Output 3

185297239

Be sure to print the output modulo 10^9 + 7.

Solution in Rust (1.15.1) (judge verdict: Runtime Error)

// Template from https://techracho.bpsinc.jp/jhonda/2019_08_05/78537

// input macro for competive-programming from 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)*}
    };
}

#[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)*}
    };

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

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

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

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

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

fn main() {
    use std::io::Read;
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();

    println!("{}", solve(&s));
}

macro_rules! test {
    ($($input:expr => $output:expr),* $(,)*) => {
        #[test]
        fn solve_test() {
            $(
                assert_eq!(solve($input), $output);
            )*
        }
    };
}

test! {
r#"3 2
1 3 2"# => "5",

r#"2 2
2 1"# => "3"
}

fn solve(src: &str) -> String {
    input! {
        source = src,
        n: i64, k: i64,
        a: [i64; n],
    }

    let mut b = vec![0; (n * k) as usize];
    for i in 0..k {
        for j in 0..n {
            // println!("{} {}", n * i + j, j);
            b[(n * i + j) as usize] = a[j as usize];
        }
    }
    // println!("{:?}", b);

    let mut ans:i64 = 0;

    for i in 0..n * k - 1 {
        for j in i+1..n * k {
            if b[i as usize] > b[j as usize] {
                ans += 1;
                // println!("{} {}", i, j);
            }
        }
    }

    ans %= 1_000_000_000 + 7;


    ans.to_string()
}