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)

n, k = gets.chomp.split(' ').map(&:to_i)
a = gets.chomp.split(' ').map(&:to_i).sort

M = (1e9+7).to_i

if a.any? { |i| i > 0 } || k % 2 == 0

  b = []
  (0...(k / 2)).each do |m|
    if a[0] * a[1] >= a[-2] * a[-1]
      b.append(a.shift(2))
    else
      b.append(a.pop(2))
    end
  end
  if k % 2 == 1
    b.append(a.pop)
  end

  answer = 1
  b.each do |i|
    answer *= i
    answer %= M
  end
  puts answer
else
  answer = 1
  a[(n - k)...n].each do |i|
    answer *= i
    answer %= M
  end
  puts answer
end