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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
fscanf(STDIN, "%d %d", $n, $k);
$a_number = explode(' ', trim(fgets(STDIN)));
$a_multi = array();
for($i = 0; $i < $n - 1; ++$i){
	$a_multi[$i] = array();
	for($h = $i + 1; $h < $n; ++$h){
		$a_multi[$i][] = $a_number[$i] * $a_number[$h];
	}
}

$ans = $a_multi[0][0];
array_shift($a_multi[0]);
--$k;
while($k-- > 0){
	if(count($a_multi) > 1){
		if($a_multi[0][0] <= $a_multi[1][0]){
			$ans = $a_multi[0][0];
			array_shift($a_multi[0]);
			continue;
		}else{
			$ans = $a_multi[1][0];
			array_shift($a_multi[1]);
			continue;
		}
	}else{
		if(count($a_multi[0]) > 1){
			$ans = $a_multi[0][0];
			array_shift($a_multi[0]);
			continue;
		}else{
			$ans =  $a_multi[0][0];
			break;
		}
	}
}

echo $ans;
?>