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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from sys import stdout
printn = lambda x: stdout.write(str(x))
inn = lambda : int(input())
inl   = lambda: list(map(int, input().split()))
inm   = lambda:      map(int, input().split())
ins = lambda : input().strip()
DBG = True  and False
BIG = 999999999
R = 10**9 + 7

def ddprint(x):
  if DBG:
    print(x)

def hasbelow1(x,b,j):
    if b==0:
        ret = 0 if x<=0 else len(a)-j
    elif b>0:
        mn = j-1
        mx = len(a)
        while mx>mn+1:
            mid = (mx+mn)//2
            if b*a[mid]<x:
                mn = mid
            else:
                mx = mid
        ret = mx-j
    else:
        mn = j-1
        mx = len(a)
        while mx>mn+1:
            mid = (mx+mn)//2
            if b*a[mid]<x:
                mx = mid
            else:
                mn = mid
        ret = len(a)-mx
    #if b!=0:
    #    ddprint("x {} b {} j {} mn {} mx {} ret {}".format(x,b,j,mn,mx,ret))
    return ret

def hasbelow(x,m):
    #ddprint("hasb x {} m {}".format(x,m))
    sm = 0
    for i in range(n-1):
        h = hasbelow1(x,a[i],i+1)
        #ddprint("i {} ai {} h {}".format(i,a[i],h))
        sm += h
        if sm>=m:
            return True
    return False

n,k = inm()
a = inl()
a.sort()
mx = 10**20
mn = -mx
while mx>mn+1:
    mid = (mx+mn)//2
    if hasbelow(mid,k):
        mx = mid
    else:
        mn = mid
    #ddprint("mx {} mn {}".format(mx,mn))
print(mn)