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)


import Data.List
import Control.Monad
import qualified Data.Vector.Unboxed as V
import qualified Data.ByteString.Char8 as BS

getIntList :: IO [Int]
getIntList = unfoldr (BS.readInt . BS.dropWhile (== ' ')) <$> BS.getLine

mod' :: Int -> Int
mod' n = if n < 1000000007 then n else n `mod` 1000000007

cnt :: Int -> Int -> Int -> Int
cnt k inv lt = mod' $ (mod' $ inv*k) + (mod' $ lt*((k*(k-1))`div`2))

main :: IO ()
main = do
  [n,k] <- getIntList
  as <- getIntList
  let vec = (V.fromList as)
      inv = mod' $ foldl' (\x y -> mod' $ x + y) 0 [if (vec V.! i) > (vec V.! j) then 1 else 0 | i <- [0..n-1], j <- [0..n-1], i < j]
      lt = mod' $ (inv+) $ mod' $  foldl' (\x y -> mod' $ x + y) 0 [if (vec V.! i) > (vec V.! j) then 1 else 0 | i <- [0..n-1], j <- [0..n-1], i > j]
  print $ cnt k inv lt