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)

MOD = 1e+9.to_i + 7

def total(n)
  (n * (n + 1)) / 2
end

def main(argv)
  (n, k) = gets.chomp.split(' ').map(&:to_i)
  a = gets.chomp.split(' ').map(&:to_i)
  
  cnt = 0
  
  n.times do |i|
    l = 0
    (0...i).each do |j|
      if a[j] < a[i] then
        l += 1
      end
    end
    
    r = 0
    ((i + 1)...n).each do |j|
      if a[j] < a[i] then
        r += 1
      end
    end
    
    xl = (total(k - 1) * l) % MOD
    cnt = (cnt + xl) % MOD
    
    xr  = (total(k) * r) % MOD
    cnt = (cnt + xr) % MOD
  end
  
  puts cnt.to_s
end

if self.to_s == 'main' then
  main(ARGV)
end