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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

base :: Int
base = 1000000007

solve :: Int -> Int -> [Int] -> Int
solve n k a = ((((n_inv * k) `mod` base) * (k + 1)) `div` 2 -  n_ordered * k) `mod` base
--solve n k a = ((n_inv * k) `div` base) * (k + 1) --`div` 2 - n_ordered * k
--solve n k a = n_inv + n_ordered
  where
    calc _ 0 _ = (0, 0)
    calc (h:rs) x y = (a0 + b0, a1 + b1)
      where
        (a0, a1) = calc rs (x - 1) y
        (b0, b1)
          | h < y = (1, 1)
          | h > y = (1, 0)
          | True  = (0, 0)
    (n_inv, n_ordered) = foldl (\(x0, y0) (x1, y1) -> (x1 + x0, y1 + y0)) (0, 0) $ zipWith (calc a) [1 .. ] $ tail a

main :: IO ()
main = do
  n:k:_ <- fmap (\x -> read x :: Int) . words <$> getLine
  a <- fmap (\x -> read x :: Int) . words <$> getLine
  print $ solve n k a