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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;

public class Main {

	static void quick_sort(long[] d, int left, int right) {
		if (left >= right) {
			return;
		}
		long p = d[(left + right) / 2];
		int l = left, r = right;
		long tmp;
		while (l <= r) {
			while (d[l] < p) {
				l++;
			}
			while (d[r] > p) {
				r--;
			}
			if (l <= r) {
				tmp = d[l];
				d[l] = d[r];
				d[r] = tmp;
				l++;
				r--;
			}
		}
		quick_sort(d, left, r); // ピボットより左側をクイックソート
		quick_sort(d, l, right); // ピボットより右側をクイックソート
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		int c = N * (N - 1) / 2;
		long set[] = new long[c];
		long number[] = new long[N];
		for (int i = 0; i < N; i++) {
			number[i] = sc.nextLong();
		}
		int k = 0;
		for (int i = 0; i < N - 1; i++) {
			for (int j = i + 1; j < N; j++) {
				if (i != j) {
					set[k] = number[i] * number[j];
					k++;
				}
			}
		}

		quick_sort(set, 0, c - 1);

		System.out.println(set[K - 1]);

	}
}