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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
define('MOD', 10 ** 9 + 7);

fscanf(STDIN, '%d %d', $n, $k);
$a = array_map('intval', explode(' ', trim(fgets(STDIN))));

$cnt1 = $cnt2 = 0;
for ($i = 0; $i < $n; $i++) {
    for ($j = 0; $j < $n; $j++) {
        if ($i === $j) continue;

        if ($i < $j && $a[$i] > $a[$j]) $cnt1++;
        elseif ($i > $j && $a[$i] > $a[$j]) $cnt2++;
    }
}

$x1 = (1 + $k) * $k / 2 % MOD;
$x2 = (1 + $k - 1) * ($k - 1) / 2 % MOD;
$ans = (($cnt1 * $x1) % MOD + ($cnt2 * $x2) % MOD) % MOD;
echo $ans . PHP_EOL;