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 Julia (1.4.0) (judge verdict: Runtime Error)

# SPDX-License-Identifier: X11
# 2020-08-30
# Multiplication 4 (500pt)

const M = 1000000007

function main()
  N, K = parse.(Int, split(readline(), " "))
  A = parse.(Int, split(readline(), " "))
  T = 1

  if N == K
    for i ∈ A
      T *= i ⨳ M
      T = T ⨳ M
    end
    println(T)
    return
  end

  sort!(A; lt = test, rev = true)

  B = A[1:K]
  if oddneg(B)
    R = A[K + 1:end]
    fp = firstpos(R) === nothing ? -1 : firstpos(R)
    fpv = fp == -1 ? fp : R[fp]
    fn = firstneg(R) === nothing ? -1 : firstneg(R)
    fnv = fn == -1 ? fn : abs(R[fn])

    if fpv > fnv || fpv < fnv && lastpos(B) === nothing && fpv > -1
      B[lastneg(B)] = R[fp]
    elseif fpv < fnv && !(lastpos(B) === nothing)
      B[lastpos(B)] = R[fn]
    elseif fpv == fnv
      B[K] = B[K] < 0 ? R[fp] : R[fn]
    else
      if A[N] == 0
        println(0)
        return
      else
        B = A[end + 1 - K:end]
      end
    end
  end

  for i ∈ B
    T *= i ⨳ M
    T = T ⨳ M
  end

  println(T)
end

function test(x::Int, y::Int)
  return abs(x) == abs(y) ? x < y : abs(x) < abs(y);
end

⨳(x, y) = mod(x, y)
oddneg(A) = count(x -> x < 0, A) & 1 == 1
firstneg(A) = findfirst(x -> x < 0, A)
firstpos(A) = findfirst(x -> x > 0, A)
lastneg(A) = findlast(x -> x < 0, A)
lastpos(A) = findlast(x -> x > 0, A)

main()