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>
#include <math.h>
#define N_MAX 210000

int Sort( int *a, int len );

int main (){
    
    //標準入力から条件を入力
    int N, K;
    int A[N_MAX];
    scanf("%d%d",&N,&K);
    for ( int i=0; i<N; i++ ){
        scanf("%d",&A[i]);
    }
    
    //multiple[]に、for二重ループにより積の値を代入していく
    int multiple[N*N/2];
    int *p = &multiple[0];
    
    for ( int i=0; i<N-1; i++ ){
        for ( int j=i+1; j<N; j++ ){
            *p = ( A[i] * A[j] );
            p++;
        }
    }
    
    //バブルソートにより小さい順に並べ替える。
    Sort( multiple, N );
    
    printf("%d",multiple[K-1]);
    return 0;
}

//*バブルソート関数*//
int Sort ( int *a, int len ){
    int tmp;
    for ( int i=0; i<len; i++ ){
        for ( int j=len-1; j>i; j-- ){
            if( a[j-1] > a[j] ){
                tmp = a[j-1];
                a[j-1] = a[j];
                a[j] = tmp;
            }
        }
    }
    return *a;
}