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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use std::mem;

fn main() {
    input! {
        n: usize,
        k: isize,
        a: [isize; n],
    }

    let mo = 1_000_000_007;
    let mut v_first = vec![0; n];
    let mut v_repeat = vec![0; n];

    for i in 0..n {
        for j in i + 1..n {
            if a[i] > a[j] {
                v_first[i] += 1;
            }
        }
    }
    for i in 0..n {
        for j in 0..n {
            if a[i] > a[j] {
                v_repeat[i] += 1;
            }
        }
    }

    let sum_first: isize = v_first.iter().sum();
    let sum_repeat: isize = v_repeat.iter().sum();
    //ans += (sum_first + (sum_repeat % mo) * (k - i - 1) % mo) % mo;
    let ans = (k * sum_first % mo + (sum_repeat * (k * (k - 1) % mo) % mo) * modinv(2, mo) % mo) % mo;
    println!("{}", ans);
}

fn modinv(mut a: isize, m: isize) -> isize {
    let mut b: isize = m;
    let mut u: isize = 1;
    let mut v = 0;

    while b != 0 {
        let t = a / b;
        a -= t * b;
        mem::swap(&mut a, &mut b);
        u -= t * v;
        mem::swap(&mut u, &mut v);
    }

    u %= m;

    if u < 0 {
        u += m;
    }
    u
}