#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
using namespace std;
const char sp = ' ';
const long long N = 2e5 + 10;
const long long INF = 2e18;
long long n;
long long k, a[N], L, R;

int main() {
	cin >> n >> k;
	for (long long i = 0; i < n; i++) {
		cin >> a[i];
	}
	sort(a, a + n);
	L = -INF;
	R = INF;
	while (L + 1 != R) {
		long long Mid = (L + R) / 2;
		long long cnt = 0;
		for (long long i = 0; i < n; i++) {
			if (a[i] < 0) {
				long long l, r;
				l = i;
				r = n;
				while (l + 1 != r) {
					long long mid = (l + r) / 2;
					if (a[i] * a[mid] <= Mid) {
						r = mid;
					}
					else {
						l = mid;
					}
				}
				cnt += n - r;
			}
			else {
				long long l, r;
				l = i;
				r = n;
				while (l + 1 != r) {
					long long mid = (l + r) / 2;
					if (a[i] * a[mid] <= Mid) {
						l = mid;
					}
					else {
						r = mid;
					}
				}
				cnt += l - i;
			}
		}
		if (cnt < k)L = Mid;
		else R = Mid;
	}
	cout << R << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

Examples

Input

4 3
3 3 -4 -2


Output

-6


Input

10 40
5 4 3 2 -1 0 0 0 0 0


Output

6


Input

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


Output

448283280358331064

### Response

```cpp
#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
using namespace std;
const char sp = ' ';
const long long N = 2e5 + 10;
const long long INF = 2e18;
long long n;
long long k, a[N], L, R;

int main() {
	cin >> n >> k;
	for (long long i = 0; i < n; i++) {
		cin >> a[i];
	}
	sort(a, a + n);
	L = -INF;
	R = INF;
	while (L + 1 != R) {
		long long Mid = (L + R) / 2;
		long long cnt = 0;
		for (long long i = 0; i < n; i++) {
			if (a[i] < 0) {
				long long l, r;
				l = i;
				r = n;
				while (l + 1 != r) {
					long long mid = (l + r) / 2;
					if (a[i] * a[mid] <= Mid) {
						r = mid;
					}
					else {
						l = mid;
					}
				}
				cnt += n - r;
			}
			else {
				long long l, r;
				l = i;
				r = n;
				while (l + 1 != r) {
					long long mid = (l + r) / 2;
					if (a[i] * a[mid] <= Mid) {
						l = mid;
					}
					else {
						r = mid;
					}
				}
				cnt += l - i;
			}
		}
		if (cnt < k)L = Mid;
		else R = Mid;
	}
	cout << R << endl;
}

```