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;

const MOD: usize = 1_000_000_007;


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

    let mut after: Vec<usize> = vec![];
    let mut before: Vec<usize> = vec![];

    for i in 0..n {
        let mut count = 0;
        for j in 0..i {
            if aa[j] < aa[i] {
                count += 1;
            }
        }
        before.push(count);

        let mut count = 0;
        for j in i+1..n {
            if aa[j] < aa[i] {
                count += 1;
            }
        }
        after.push(count);
    }
    dbg!(&before, &after);
    for b in before.iter_mut() {
        if k % 2 == 0 {
            *b *= k / 2;
            *b %= MOD;
            *b *= k - 1;
            *b %= MOD;
        } else {
            *b *= k;
            *b %= MOD;
            *b *= (k - 1) / 2;
            *b %= MOD;
        }
    }
    for a in after.iter_mut() {
        let temp = *a * k;
        if k % 2 == 0 {
            *a *= k / 2;
            *a %= MOD;
            *a *= k - 1;
            *a %= MOD;
        } else {
            *a *= k;
            *a %= MOD;
            *a *= (k - 1) / 2;
            *a %= MOD;
        }
        *a += temp;
        *a %= MOD;
    }

    let mut count = 0;
    for (a, b) in after.into_iter().zip(before.into_iter()) {
        count += a;
        count += b;
        count %= MOD;
    }
    println!("{}", count);
}