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 Python (3.4.3) (judge verdict: Runtime Error)

n, k = map(int, input().split())
a = list(map(int, input().split()))
po = []
ng = []
zr = 0
for i in range(n):
  if a[i]>0:
    po.append(a[i])
  elif a[i]<0:
    ng.append(a[i])
  elif a[i]==0:
    zr += 1
lp=len(po)
ln=len(ng)
negn=lp*ln
zern=zr*(n-zr)+zr*(zr-1)//2

if k <= negn:
  p=[0]*negn
  c=0
  for i in range(lp):
    for j in range(ln):
      p[c]=po[i]*ng[j]
      c+=1
  p.sort()
  print(p[k-1])
elif negn < k <= negn+zern:
  print(0)
else:
  p=[0]*(n*(n-1)//2-negn-zern)
  c=0
  for i in range(lp-1):
    for j in range(i+1, lp):
      p[c]=po[i]*po[j]
      c+=1
  for i in range(ln-1):
    for j in range(i+1, ln):
      p[c]=ng[i]*ng[j]
      c+=1
  p.sort()
  print(p[k-1-negn-zern])