//its very good question i think binary search conceptual.................................................................................
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int n;
ll k, a[202020];
ll st = -2e18, en = 2e18;

int main()
{
	cin >> n >> k;
	for (int i = 0; i < n; i++) cin >> a[i];
	sort(a, a + n);
	while (st + 1 != en)
	{
		ll mid = (st + en) / 2, cnt = 0;
		for (int i = 0; i < n; i++)
		{
			if (a[i] < 0)
			{
				int l = i, r = n;
				while (l + 1 != r)
				{
					int m = (l + r) / 2;
					if (a[i]*a[m] <= mid) r = m;
					else l = m;
				}
				cnt += n - r;
			}
			else
			{
				int l = i, r = n;
				while (l + 1 != r)
				{
					int m = (l + r) / 2;
					if (a[i]*a[m] <= mid) l = m;
					else r = m;
				}
				cnt += l - i;
			}
		}
		if (cnt < k) st = mid;
		else      en = mid;
	}
	cout << en << endl;
	return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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
//its very good question i think binary search conceptual.................................................................................
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int n;
ll k, a[202020];
ll st = -2e18, en = 2e18;

int main()
{
	cin >> n >> k;
	for (int i = 0; i < n; i++) cin >> a[i];
	sort(a, a + n);
	while (st + 1 != en)
	{
		ll mid = (st + en) / 2, cnt = 0;
		for (int i = 0; i < n; i++)
		{
			if (a[i] < 0)
			{
				int l = i, r = n;
				while (l + 1 != r)
				{
					int m = (l + r) / 2;
					if (a[i]*a[m] <= mid) r = m;
					else l = m;
				}
				cnt += n - r;
			}
			else
			{
				int l = i, r = n;
				while (l + 1 != r)
				{
					int m = (l + r) / 2;
					if (a[i]*a[m] <= mid) l = m;
					else r = m;
				}
				cnt += l - i;
			}
		}
		if (cnt < k) st = mid;
		else      en = mid;
	}
	cout << en << endl;
	return 0;
}

```