Problem:
Score : 500 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.

Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Constraints

1 \leq K \leq N \leq 2\times 10^5

|A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 \ldots A_N

Output

Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Sample Input 1

4 2
1 2 -3 -4

Sample Output 1

12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

Sample Input 2

4 3
-1 -2 -3 -4

Sample Output 2

1000000001

The possible products of the three chosen elements are -24, -12, -8, and -6, so the maximum product is -6.

We print this value modulo (10^9+7), that is, 1000000001.

Sample Input 3

2 1
-1 1000000000

Sample Output 3

1000000000

The possible products of the one chosen element are -1 and 1000000000, so the maximum product is 1000000000.

Sample Input 4

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1

Sample Output 4

999983200

Be sure to print the product modulo (10^9+7).

Solution in Haskell (GHC 8.8.3) (judge verdict: Wrong Answer)

import qualified Data.ByteString.Char8 as BS
import Data.Function
import Data.List
import Data.Maybe
import Data.Ord

readInt = fst . fromJust . BS.readInt

readIntList = map readInt . BS.words

getIntList = readIntList <$> BS.getLine

prod = foldl1' (\x y -> mod1G7 (x * y))

mod1G7 n = n `mod` (10 ^ 9 + 7)

main :: IO ()
main = do
  [n, k] <- getIntList
  as <- getIntList
  if k == n
    then print $ prod as
    else
      if all (<= 0) as
        then
          if odd k
            then print $ prod (take k (sortBy (compare `on` Down) as))
            else print $ prod (take k (sort as))
        else do
          let as' = sortBy (comparing (\a -> (- (abs a)))) as
          let bs = take k as'
          if elem 0 bs
            then print 0
            else
              if even (length (filter (\x -> x < 0) bs))
                then print $ prod bs
                else do
                  let bsr = reverse bs
                  let asr = drop k as'
                  let bs' = filter (\b' -> b' < 0) bsr
                  let as'' = filter (\a -> a >= 0) asr
                  let c = if length as'' == 0 || length bs' == 0 then 0 else prod ((head as'') : (delete (head bs') bs))
                  let bs'' = filter (\b' -> b' > 0) bsr
                  let as''' = filter (\a -> a <= 0) asr
                  let c' = if length as''' == 0 || length bs'' == 0 then 0 else prod ((head as''') : (delete (head bs'') bs))
                  if c == 0 || c' == 0
                    then
                      if c == 0 && c' == 0
                        then print 0
                        else print $ maximum [c, c']
                    else do
                      let rat = abs ((fromIntegral (head as'')) / (fromIntegral (head (bs'))))
                      let rat' = abs ((fromIntegral (head as''')) / (fromIntegral (head (bs''))))
                      if rat > rat'
                        then print c
                        else print c'