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)

N, K = gets.split.map(&:to_i)
ary = gets.split.map(&:to_i).sort
if K.odd? && ary[-1] < 0
    ary.reverse!
else
    ary.sort!{|a, b| b.abs <=> a.abs}
end
sum = 1
last_minus_idx = K
last_plus_idx = K
K.times do |i|
    sum = sum * ary[i] % (10**9 + 7)
    if ary[i] < 0
        last_minus_idx = i
    else
        last_plus_idx = i
    end
end
if K < N && sum < 0
    sum1, sum2 = sum, sum
    # マイナスの入れ替え
    target = ary[K..-1].find{|i| i > 0}
    if target
        sum1 = sum1 / ary[last_minus_idx] * target
    end
    # プラスの入れ替え
    target = ary[K..-1].find{|i| i < 0}
    if target
        sum2 = sum2 / ary[last_plus_idx] * target
    end
    # sum1とsum2の比較
    sum = [sum1, sum2].max
end
puts sum