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: Time Limit Exceeded)

N,K = list(map(int,input().split()))
A = list(map(int,input().split()))

pos = []
neg = []
zero = []

A.sort()

A2 = [A[i]**2 for i in range(N)] 
A2.sort()
for i in range(N):
    if A[i] > 0:
        pos.append(A[i])
    elif A[i] == 0:
        zero.append(A[i])
    else:
        neg.append(A[i])
      
import bisect


#条件に合うものを重複して数えて後から重複しない組み合わせに変換
def fun(x):
    count = 0
    #xが0より大きいとき、後ろにある0は何個あるか
    if x >= 0:
        count += len(zero)*N
    #xを正で割ったとき、Aの中に何個それより大きいものがあるか
    for i in range(len(pos)):
        count += bisect.bisect_right(A,x//pos[i])
    #xを負で割ったとき、Aの中に何個それより小さいものがあるか(符合が逆転)
    #x+1にする(同じ数字がAにあるから)
    for i in range(len(neg)):
        count += N - bisect.bisect_right(A,(x+1)//neg[i])
    #条件に合う同じ要素を選ぶものを除く    
    count -= bisect.bisect_right(A2,x)
    #半分にすれば重複しない組み合わせになる        
    return count//2

left = -10**18
right = 10**18

#二分探索する
while right - left > 1:
    mid = (left+right)//2
    
    if fun(mid) >= K:
        right = mid
    else:
        left = mid
        
print(right)