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 (2.7.1) (judge verdict: Accepted)

d = 10**9+7
n, k = gets.split.map(&:to_i)
a = gets.split.map(&:to_i)
x = 0
(0..n-1).each do |i|
  (i+1..n-1).each do |j|
    x += 1 if a[i] > a[j]
  end
end
a = a.sort.reverse
y = 0
(0..n-1).each do |i|
  (i+1..n-1).each do |j|
    y += 1 if a[i] > a[j]
  end
end
ans = (x * k) % d
s = (k * k) % d
t = (k * (k + 1) / 2) % d
z = y * (s - t) % d
ans += z
puts ans % d