Problem:
Score: 400 points

Problem Statement

We have N integers A_1, A_2, ..., A_N.

There are \frac{N(N-1)}{2} ways to choose two of them and form a pair. If we compute the product of each of those pairs and sort the results in ascending order, what will be the K-th number in that list?

Constraints

All values in input are integers.

2 \leq N \leq 2 \times 10^5

1 \leq K \leq \frac{N(N-1)}{2}

-10^9 \leq A_i \leq 10^9\ (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \dots A_N

Output

Print the answer.

Sample Input 1

4 3
3 3 -4 -2

Sample Output 1

-6

There are six ways to form a pair. The products of those pairs are 9, -12, -6, -12, -6, 8.

Sorting those numbers in ascending order, we have -12, -12, -6, -6, 8, 9. The third number in this list is -6.

Sample Input 2

10 40
5 4 3 2 -1 0 0 0 0 0

Sample Output 2

6

Sample Input 3

30 413
-170202098 -268409015 537203564 983211703 21608710 -443999067 -937727165 -97596546 -372334013 398994917 -972141167 798607104 -949068442 -959948616 37909651 0 886627544 -20098238 0 -948955241 0 -214720580 277222296 -18897162 834475626 0 -425610555 110117526 663621752 0

Sample Output 3

448283280358331064

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

N, K = gets.chomp.split(" ").map(&:to_i)
A = gets.chomp.split(" ").map(&:to_i)

pogi = []
nega = []
zero = 0

h = A.group_by {|a| a <=> 0} # 返り値はハッシュ
nega = h[-1] || [] # ハッシュから配列を取り出す
pogi = h[1] || []
zero = h[0] || []

pogi.sort!
nega.sort!.reverse!

n_size = nega.size
p_size = pogi.size
z_size = zero.size

pair_zero = z_size * (z_size - 1) / 2 + z_size * (p_size + n_size)
pair_nega = p_size * n_size
pair_pogi = p_size * (p_size - 1) + n_size * (n_size - 1)

if pair_nega < K && K <= (pair pair_nega + pair_zero)
    puts 0 
    exit
end

    # 2分木探査の返り値 b を res に入れる

res = if K <= pair_nega
    (-10**19..0).bsearch do |b|
        sum = 0
        r = p_size - 1
        nega.each do |n|
            while r >= 0 && n * pogi[r] <= b
                r -= 1
            end
            sum += p_size - r - 1
        end
        sum >= K
    end
else
    (0..10**19).bsearch do |b|
        sum = 0
        [nega, pogi].each do |a| # 負の数同士、正の数同士考える
            l = 0
            r = a.size - 1　
            while l < r
                while l < r && a[l] * a[r] > b　# 最小の値　×　最大の値
                    r -= 1
                end
                sum += r - l
                l += 1
            end
        end
        sum >= K - pair_nega - pair_zero
    end
end

puts res