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: Accepted)

require 'set'; require 'prime'; require 'pp'
INF=Float::INFINITY; MOD=10**9+7
n,k = gets.chomp.split.map(&:to_i)
a = gets.chomp.split.map(&:to_i)
a.sort!

plusc = a.count{|e| e > 0}
minusc = a.count{|e| e < 0}
has_zero = a.any?{|e| e == 0}

if plusc == 0 && k.odd?
  if has_zero
    puts 0
  else
    ans = 1
    k.times do |i|
      ans = ans * a[-i-1] % MOD
      # ans = ans % MOD if i.even?
    end
    ans = (ans + MOD) % MOD
    puts ans
  end
  exit
end
# -------


ans = 1
l = 0
r = n - 1

if k.even?
  while l + (n-1-r) + 2 <= k
    lmul = a[l] * a[l+1]
    rmul = a[r-1] * a[r]
    if lmul > rmul || (plusc <= 2 && k.odd?)
      ans = ans * lmul % MOD
      l += 2
      minusc -= 2
    else
      ans = ans * rmul % MOD
      r -= 2
      plusc -= 2
    end
    # p [[l, r], [lmul, rmul], ans]
  end

else
  while l + (n-1-r) + 2 <= k
    lmul = a[l] * a[l+1]
    rmul = a[r-1] * a[r]
    if lmul * a[r] > rmul * a[r-2] || (plusc <= 2 && k.odd?)
      ans = ans * lmul % MOD
      l += 2
      minusc -= 2
    else
      ans = ans * rmul % MOD
      r -= 2
      plusc -= 2
    end
    # p [[l, r], [lmul, rmul], ans]
  end
end


if l + (n-1-r) < k
  ans = [ans * a[l], ans * a[r]].max
end

ans = (ans + MOD) % MOD
puts ans


# 0の場合！！！