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: Accepted)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

#[allow(dead_code)]
fn with_stdout_lock<F: FnOnce(std::io::BufWriter<std::io::StdoutLock>)>(f: F) {
    let out = std::io::stdout();
    let w = std::io::BufWriter::new(out.lock());
    f(w);
}

#[allow(unused_macros)]
macro_rules! debug {
    ($var:expr) => {
        match &$var { v => eprintln!(concat!(stringify!($var), "={:?}"), v) }
    };
    ($var:expr, $($vars:expr),*) => {
        match &$var { v => eprint!(concat!(stringify!($var), "={:?} "), v) }
        debug!($($vars),*);
    };
}

fn main() {
    let n: usize = input();
    let k: i64 = input();
    let a: Vec<i32> = (0..n).map(|_| input()).collect();
    const MOD: i64 = 1000000007;

    let mut t = 0;
    for i in 0..n {
        for j in i + 1..n {
            if a[i] > a[j] {
                t += 1;
            }
        }
    }

    let mut ans = t * k as i64 % MOD;
    for i in 0..n {
        let mut c = 0;
        for j in 0..n {
            if a[i] > a[j] {
                c += 1;
            }
        }
        ans += k * (k - 1) / 2 % MOD * c % MOD;
        ans %= MOD;
    }

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