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)

N, K = gets.split(' ').map(&:to_i)
A = gets.split(' ').map(&:to_i)
r = 1000 * 1000 * 1000 + 7
A_MAX = 2000

tento = (0..A.size).to_a.map { |i| Array(A[i+1..-1]).select { |n| n < A[i] }.size }.reduce(:+)
cnt = [0] * (A_MAX + 1)
A.each do |n|
  cnt[n] += 1
end

belows = 0
aboves = 0
(1...A_MAX).each do |n|
  belows += cnt[n - 1]
  aboves += (belows * cnt[n]) % r
  # pp [n, belows, aboves]
end

betweens = (K - 1) * K / 2

hoge = aboves * betweens % r

puts ((tento * K % r) +  hoge ) % r