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 C (Clang 3.8.0) (judge verdict: Runtime Error)

#include<stdio.h>
int main()
{
    int a,c,d,e=0,f,b;
    scanf("%d %d",&a,&b);
    long long int  op[a],od[a*(a-1)/2],ob;
    for(c=0;c<a;c++)
        scanf("%l64d",&op[c]);
    for(c=0;c<a-1;c++)
    {
        for(d=c+1;d<a;d++)
        { if(op[d]==0||op[c]==0) od[e]=0; else od[e]=op[c]*op[d]; e++;}
    }
    for(c=0;c<e;c++)
    {   ob=od[c]; f=c;
        for(d=c+1;d<e;d++)
        if(ob>od[d]) {ob=od[d]; f=d; }
        od[f]=od[c];
        od[c]=ob;
    }
    printf("%l64d",od[b-1]);
    return 0;
}