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 Fortran (gfortran v4.8.4) (judge verdict: Runtime Error)

program main

  implicit none

  integer    :: n,k
  integer,allocatable :: a(:)

  integer    :: i,j,pn
  integer(8),allocatable :: p(:)

  read(*,*) n, k
  allocate( a(n) )
  !allocate( p(int(n*(n-1))+1) )
  allocate( p(n*(n-1)/2) )
  read(*,*) a
  
  pn = 0
  do i = 1, n
    do j = i+1, n
      pn = pn + 1
      p(pn) = a(i) * a(j)
    end do
  end do
  
  !call quicksort1( p, 1, int( n*(n-1) )+1 ) 
  call quicksort1( p, 1, n*(n-1)/2 ) 

  print*, p(k)

end program main

recursive subroutine quicksort1(a, first, last)
  implicit none
  integer(8),intent(inout)::a(*)
  integer,intent(in)::first,last
  !
  ! Original
  ! https://gist.github.com/t-nissie/479f0f16966925fa29ea
  !
  integer(8)::x,t
  integer::i,j

  x = a((first+last)/2)
  i = first
  j = last
  do
     do while (a(i) < x)
        i=i+1
     end do
     do while (x < a(j))
        j=j-1
     end do
     if (i >= j) exit
     t = a(i)
     a(i) = a(j)
     a(j) = t
     i=i+1
     j=j-1
  end do
  if (first < i-1) call quicksort1(a, first, i-1)
  if (j+1 < last)  call quicksort1(a, j+1, last)

  return
end subroutine quicksort1