p02774 AtCoder Beginner Contest 155 - Pairs
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.

Examples

Input

4 3
3 3 -4 -2


Output

-6


Input

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


Output

6


Input

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


Output

448283280358331064
```python
N,K,*A = map(int, open(0).read().split())
Ap = []
An = []
Az = []
for c in A:
  if c>0:
    Ap.append(c)
  elif c<0:
    An.append(c)
  else:
    Az.append(c)
l = -10**20
r = 10**20
Ap.sort(reverse=True)
An.sort()
while l+1<r:
  s = (l+r)//2
  cnt = 0
  j = len(Ap)-1
  for i in range(len(Ap)):
    while j>-1 and Ap[i]*Ap[j]<s:
      j -= 1
    cnt += len(Ap)-1-j
  for i in range(len(Ap)):
    if Ap[i]*Ap[i]<s:
      cnt -= 1
  j = len(An)-1
  for i in range(len(An)):
    while j>-1 and An[i]*An[j]<s:
      j -= 1
    cnt += len(An)-1-j
  for i in range(len(An)):
    if An[i]*An[i]<s:
      cnt -= 1
  cnt //= 2
  if s>0:
    cnt += len(Az)*(len(Az)-1)//2 + len(Az)*(len(Ap)+len(An)) + len(Ap)*len(An)
  else:
    j = 0
    for i in range(len(Ap)-1,-1,-1):
      while j<len(An) and Ap[i]*An[j]<s:
        j += 1
      cnt += j
  if cnt>=K:
    r = s
  else:
    l = s
print(l)

```