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 Ruby (2.7.1) (judge verdict: Wrong Answer)

# frozen_string_literal: true

n, k = gets.split.map(&:to_i)
a = gets.split.map(&:to_i)
MOD = 10**9 + 7
def prod(a)
  p = 1
  a.each do |e|
    p *= e
    p %= MOD
  end
  p
end
aa = a.sort_by(&:abs).reverse

b = aa.take(k)

max = 0
if b.any?(0)
  max = 0
elsif b.count(&:negative?).even? || n == k
  max = prod(b)
else
  c = aa[k..-1]
  mbi = []
  if c.any?(&:positive?)
    m = b.filter(&:negative?).max
    i = b.index(m)
    b[i] = c.max
    max = prod(b)
    mbi = [m, b[i]]
  end
  if b.any?(&:positive?)
    m = b.filter(&:positive?).min
    i = b.index(m)
    b[i] = c.min
    if mbi.empty?
      max = prod(b)
    elsif (mbi[1] * m).abs < (mbi[0] * b[i]).abs
      max = prod(b)
    end
  elsif a.any?(0)
    max = 0
  else
    max = prod(a.sort.reverse.take(k))
  end
end
puts max