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 Ruby (2.3.3) (judge verdict: Wrong Answer)

require 'pp'

MOD = 10 ** 9 + 7

n, k = gets.chomp!.split(" ").map(&:to_i)
a_n = gets.chomp!.split(" ").map(&:to_i)

ans = Array.new(n, 0)
ans2 = Array.new(n, 0)

(n-1).times do |i|
    (i+1).upto(n-1) do |j|
        if a_n[i] > a_n[j]
            ans[i] += 1
        end
    end
end

sorta_n = a_n.sort!.reverse!
(n-1).times do |i|
    (i+1).upto(n-1) do |j|
        if sorta_n[i] > sorta_n[j]
            ans2[i] += 1
        end
    end
end

def ss(a, d, n)
    (n * (2 * a + (n - 1) * d)) / 2
end

sum = 0
ans2.each do |i|
    sum = (sum + ss(i, i, k - 1)) % MOD
end

ans.length.times do |i|
    sum = (ans[i] + sum + ans2[i] * (k - 1)) % MOD
end

pp sum % MOD