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 Data.List

headc [] = 1
headc (s:ss) = s

solve :: [Integer] -> [Integer] -> Int -> Integer
solve [] [] 0 = 1
solve (p:ps) (m:ms) 0 = 1
solve (p:ps) [] 0 = 1
solve [] (m:ms) 0 = 1
solve (p:ps) [] i = p * solve ps [] (i-1)
solve [] (m:ms) i = m * solve [] ms (i-1)
solve (p:ps) (m:ms) i | p >= mm = p * solve ps (m:ms) (i-1)
                      | i >= 2 = mm * solve (p:ps) (tail ms) (i-2)
                      | otherwise = 1
                        where
                            mm = m * (headc ms)

main = do
    [n,k] <- map read . words <$> getLine :: IO [Int]
    as <- sort . map read . words <$> getLine :: IO [Integer]
    let pas = reverse $ filter (>=0) (drop ((length as) - k) as)
    let mas = filter (<0) (take k as)
    let msas = reverse $ filter (<0) (as)
    let scs = solve pas mas k
    let mis = solve pas msas k
    if scs >= 0
        then print $ scs `mod` (1000000007)
        else print $ mis `mod` (1000000007)