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: Wrong Answer)

n, k = map(int, input().split())
A = sorted(list(map(int, input().split())))
def countA(x):
    S, T = 0, 0
    for i in range(n):
        l, r = -1, n
        if A[i] > 0:
            while r - l > 1:
                mid = (r + l) // 2
                if A[i] * A[mid] <= x:
                    l = mid
                else:
                    r = mid
            S += l + 1
        elif A[i] < 0:
            while r - l > 1:
                mid = (r + l) // 2
                if A[i] * A[mid] <= x:
                    r = mid
                else:
                    l = mid
            S += n - r
        else: # A[i]==0
            if x >= 0:
                S += n
        if A[i] ** 2 <= x:
            T += 1
    return S/2 + T

left, right = -A[-1]**2//2, A[-1]**2//2
while right - left > 1:
    mid = (right+left) // 2
    if countA(mid) >= k:
        right = mid
    else:
        left = mid
print(right)