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: Wrong Answer)

#[allow(unused_macros)]
macro_rules! input {
    ( $($t:ty),* ) => {{
        let mut s = String::new();
        std::io::stdin().read_line(&mut s);
        let mut splits = s.trim().split_whitespace();
        ($( { splits.next().unwrap().parse::<$t>().unwrap() },)*)
    }}
}

#[allow(unused_macros)]
macro_rules! invec {
    ( $ t : ty ) => {{
        let mut s = String::new();
        match std::io::stdin().read_line(&mut s) {
            Ok(0) => Vec::<$t>::new(),
            Ok(n) => s
                .trim()
                .split_whitespace()
                .map(|s| s.parse::<$t>().unwrap())
                .collect::<Vec<$t>>(),
            Err(_) => Vec::<$t>::new(),
        }
    }};
}

use std::fmt::Display;
pub fn print_vec<T: Display>(v: &Vec<T>, prefix: &str, suffix: &str) {
    if v.is_empty() {
        println!("[]");
        return;
    }
    print!("{}", prefix);
    print!("{}", v[0]);
    for i in 1..v.len() {
        print!(" {}", v[i]);
    }
    println!("{}", suffix);
}

#[allow(unused_must_use)]
#[allow(unused_variables)]
fn solve() {
    let (n, k) = input!(usize, u64);
    let a = invec!(u16);
    let m = 1_000_000_007;

    let mut inv = vec![0_u64; n];
    let mut inv_first = vec![0_u64; n];

    for i in 0..n {
        for v2 in &a {
            if a[i] > *v2 {
                inv[i] += 1;
            }
        }
    }

    for i in 0..n {
        for j in (i+1)..n {
            if a[i] > a[j] {
                inv_first[i] += 1;
            }
        }
    }

    let mut count = 0;
    let total = (k as u64 - 1) * (k as u64) / 2;
    for i in 0..n {
        if inv[i] == 0 {
            continue;
        }
        count += (total * inv[i]) % m;
    }
    

    for i in 0..(n - 1) {
        count += (inv_first[i] * k) % m;
    }


    println!("{}", count % m);
}

fn main() {
    solve();
}