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)

N, K = gets.chomp.split.map(&:to_i)
A = gets.chomp.split.map(&:to_i)
MOD = 10 ** 9 + 7

# p N, K
# p A
MAX_SIZE = 2005
d = Array.new(MAX_SIZE)
d.each.with_index do |_, i|
  d[i] = Array.new(MAX_SIZE, 0)
end

A.each.with_index do |a, i|
  1.upto(MAX_SIZE-1).each do |j|  # d[i]d[0] is ignored. They are always 0.
    if j < a  # j is larger than a
      d[i+1][j] = d[i][j] + 1
    else
      d[i+1][j] = d[i][j]
    end
  end
end

r = 0

# We use d[A.size] for the loop of k. If k = 2, then (d[i][i] + d[A.size][i]) should be added.
# If k, then (d[i][i] + (k-1) * d[A.size][i])) should be added.
# 0 + 1 + 2 + ... + (k-1) = k * (k-1) / 2

A.each.with_index do |a, i|
  # For Debug
  # p "d[last][#{a}]: #{d[A.size][a]}"
  r += (K * d[i][a]) % MOD
  r += (d[A.size][a] * K * (K-1) / 2) % MOD
  r = r % MOD
end

p r