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: Wrong Answer)

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

P = []
M = []
Z = []

A.each do |a|
  if a == 0
    Z << a
  elsif a < 0
    M << a
  else
    P << a
  end
end

ZS = Z.size
PS = P.size
MS = M.size
ZC = if ZS > 1
       Z.size * (PS + MS) + (1...ZS).inject(:+)
     else
       Z.size * (PS + MS)
     end

MC = PS * MS

# p MC
# p P
# p M

if K <= MC
  P.reverse!

  def f(x)
    pi = 1
    mi = 1

    while (pi < PS || mi < MS) && (P[pi - 1] * M[mi - 1] < x)
      if pi == PS
        mi += 1
      elsif mi == MS
        pi += 1
      elsif PS[pi - 1] < MS[mi - 1].abs
        mi += 1
      else
        pi += 1
      end
    end

    (pi * mi) >= K && P[pi - 1] * M[mi - 1] <= x
  end

  ok = 0
  ng = -1 * (10 ** 19)

  while (ok - ng).abs > 1
    x = (ok + ng) / 2

    if f(x)
      ok = x
    else
      ng = x
    end
  end

  puts x
elsif K <= MC + ZC
  puts 0
else
  L = [2, K - (MC + ZC)].max

  def f(x, nums)
    n = nums.size
    return 0 if n <= 1

    cnt = 0
    i = 0
    j = n - 1

    while i < n && j >= 0
      while j > 0 && nums[i] * nums[j] > x - 1
        j -= 1
      end

      if i <= j
        cnt += j
      else
        cnt += (j + 1)
      end

      i += 1
    end

    cnt / 2
  end

  ng = 10 ** 19
  ok = 0

  M.reverse!

  while (ok - ng).abs > 1
    x = (ok + ng) / 2
    cnt = f(x, P) + f(x, M)

    if cnt < L
      ok = x
    else
      ng = x
    end
  end

  puts ok
end