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 (GCC 5.4.1) (judge verdict: Runtime Error)

 #include <stdio.h>
 int main (void){
int K,N,x;
long int a[1000000],b[1000000];
scanf("%d%d",&N,&K);
for(int i=0;i<N;i++){
scanf("%d",a[i]);
}
for(int k=0;k<N*(N-1)/2;k++){
    for(int j=0;j<N;j++){
        for(int l=j+1;j<N;j++){
            b[k] = a[j]*a[l];
        }
    }
}
for(int u=0;u<N*(N-1)/2;u++){
    if(a[u]>a[u+1]){
        x=a[u];
        a[u] = a[u+1];
        a[u+1] = x;
    }
}
printf("%d",a[K-1]);
 }