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: Accepted)

require 'pp'
n, k = gets.split.map &:to_i
a = gets.split.map(&:to_i)

sankaku_cnt = 0

sikaku_cnt = 0

a.size.times do |i|
  a.size.times do |ii|
    sikaku_cnt += 1 if a[ii] > a[i]
  end
end
# p sikaku_cnt

x = 1
n.times do |ii|
  x.upto(n-1) do |i|
    sankaku_cnt += 1 if a[ii] > a[i]
  end
  x += 1
end
# p sankaku_cnt

res = sankaku_cnt * k + (((k-1) * k)/2) * sikaku_cnt
p res % (10**9 + 7)
# 3 0 185297239
#     185297239