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)

MOD = 10**9+7

N,K = gets.split.map &:to_i
A = gets.split.map(&:to_i).sort_by{_1.abs}

if K%2==1 && A.all?{_1<0} then
    p A[0..K-1].inject(1){|r,i| (r*i)%MOD}
    exit
end

A.reverse!
neg_count = A[0..K-1].count{_1<0}

if neg_count%2 == 1 then
    last_neg_idx = A[0..K-1].rindex{_1<0}
    last_pos_idx =  A[0..K-1].rindex{_1>=0}

    next_neg_idx = A[K..-1].index{_1<0}
    next_pos_idx = A[K..-1].index{_1>=0}

    next_neg_idx += K if next_neg_idx
    next_pos_idx += K if next_pos_idx

    swap_neg = next_pos_idx && A[next_pos_idx]/A[last_neg_idx].abs
    swap_pos = next_neg_idx && last_pos_idx && A[next_neg_idx].abs/A[last_pos_idx]
    neg = (swap_neg && swap_pos) ? swap_neg >= swap_pos : swap_neg

    if neg then
        A[last_neg_idx] = A[next_pos_idx]
    else
        A[last_pos_idx] = A[next_neg_idx]
    end
end

 p A[0..K-1].inject(1){|r,i| (r*i)%MOD}