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: Time Limit Exceeded)

<?php
// $n = (int)trim(fgets(STDIN));
$input = explode(' ', trim(fgets(STDIN)));

$n = (int)$input[0];
$k = (int)$input[1];

$input = explode(' ', trim(fgets(STDIN)));
$plus  = [];
$minus = [];
$zero  = [];
foreach ($input as $v) {
	$v = (int)$v;
	if ($v == 0) {
		$zero[] = $v;
	} elseif ($v > 0) {
		$plus[] = $v;
	} else {
		$minus[] = $v;
	}
}

// マイナスの数
$total_minus = count($minus) * count($plus);
// ゼロの数
$total_zero  = count($zero) * (count($plus) + count($minus)) + (int)floor(count($zero) * (count($zero) - 1) / 2);
// プラスの数
$total_plus  = (int)floor(count($plus) * (count($plus) - 1) / 2) + (int)floor(count($minus) * (count($minus) - 1) / 2);

if ($total_minus >= $k) {
	$all = [];
	sort($minus);
	rsort($plus);
	foreach ($minus as $m) {
		foreach ($plus as $p) {
			$all[] = $m * $p;
		}
	}
	sort($all);
	echo $all[$k-1];
} elseif ($total_minus + $total_zero >= $k) {
	echo '0';
} else {
	$all = [];
	for ($i=0; $i < count($minus); $i++) { 
		for ($j=$i+1; $j < count($minus); $j++) { 
			$all[] = $minus[$i] * $minus[$j];
		}
	}
	for ($i=0; $i < count($plus); $i++) { 
		for ($j=$i+1; $j < count($plus); $j++) { 
			$all[] = $plus[$i] * $plus[$j];
		}
	}
	sort($all);
	echo $all[$k - $total_minus - $total_zero - 1];
}