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: Runtime Error)

#!/usr/bin/env ruby

require 'set'

class Hash
  def push(key, value)
    self[key] = [] if self[key] == nil
    self[key].push(value)
  end
end

class Array
  def lower_bound(value)
    left = -1;
    right = self.length;
    while left + 1 < right
      mid = left + (right - left) / 2;
      if self[mid] >= value
        right = mid
      else
        left = mid
      end
    end
    right
  end

  def unique
    res = [self.first]
    each_cons(2) do |a, b|
      if a != b
        res.push(b)
      end
    end
    res
  end
end

def get_ints
  gets.chomp.split.map(&:to_i)
end

def get_ints_minus_one
  get_ints.map { |x| x - 1 }
end

def get_int
  gets.chomp.to_i
end

# mod m での aの逆元
def modinv(a, m)
  b = m; u = 1; v = 0

  while b != 0
    t = a / b
    a -= t * b
    c = a; a = b; b = c
    u -= t * v
    c = u; u = v; v = c
  end
  u %= m
  if u < 0
    u += m
  else
    u
  end
end

N, K = get_ints
A = get_ints

MOD = 10 ** 9 + 7

ax = A.filter { |a| a != 0 }.sort_by { |x| x.abs  }.reverse

if ax.length < K
  puts 0
else
  sum = 1
  last_minus = nil
  last_plus = nil
  ax[0...K].each do |a|
    kake = sum * a
    last_plus = a if a > 0
    last_minus = a if a < 0
    sum = (sum * a).abs % MOD
    sum = - sum if kake < 0
  end

  if sum < 0
    sum_abs = sum.abs
    # K以降に負があるかをチェック
    fu = ax[K..].find { |x| x < 0 }
    # K以降に負しかないかをチェック
    fu_only = ax[K..].all? { |x| x < 0 }


    if fu.nil?
      minv = modinv(-last_minus, MOD)
      sum = (sum_abs * minv) % MOD # 負の最後の要素を割る
      sum *= ax[K]
    else
      kouho1 = -10**9
      if last_minus
        minv = modinv(-last_minus, MOD)
        kouho1 = (sum_abs * minv) % MOD # 負の最後の要素を割る
        first_plus = ax[K..].find { |x| x > 0 }
        kouho1 *= kouho1 * first_plus if first_plus
      end

      kouho2 = -10**9
      if last_plus
        minv2 = modinv(last_plus, MOD)
        kouho2 = (sum_abs * minv2) % MOD # 正の最後の要素を割る
        first_minus = ax[K..].find { |x| x < 0 }
        kouho2 *= (first_minus).abs if first_minus
      end

      sum = [kouho1, kouho2].max
      sum %= MOD
    end
  end

  puts sum
end