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.split.map(&:to_i)
as = gets.split.map(&:to_i)
MOD = 10 ** 9 + 7

def mod_pow(x, n, mod)
  return 1 if n == 0
  res = mod_pow(x * x % mod, n / 2, mod)
  if n[0] == 1
    res = res * x % mod
  end
  res
end

count1 = 0
(0...N).each do |i|
  n = as[i]
  (0...N).each do |j|
    count1 += 1 if n > as[j]
  end
end
count1 = count1 % MOD
count1 = (count1 * (K * (K - 1) / 2 % MOD)) % MOD if K > 1

count2 = 0
(0...N).each do |i|
  n = as[i]
  ((i + 1)...N).each do |j|
    count2 += 1 if n > as[j]
  end
end

ans = (count1 + (count2 * K % MOD)) % MOD
puts ans