#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 2e5 + 2;
long long a[N];
long long n, m;
typedef long long LL;
bool check(LL x)
{
	long long h1, h2, cnt = 0;
	for (int i = 1; i <= n; i++)
	{
		if (a[i])
		{
			h1 = h2 = x / a[i];
			if (x % a[i])
			{
				if ((x < 0 && a[i]>0) || (x > 0 && a[i] < 0)) --h1;//不能直接相乘判断正负，因为会 爆
				else ++h2;
			}
			if (a[i] < 0)  cnt += n + 1 - (upper_bound(a + 1, a + 1 + n, h1) - a);
			if (a[i] > 0) cnt += lower_bound(a + 1, a + 1 + n, h2) - a - 1;
			if (a[i] * a[i] < x) cnt--;
		}
		else if (x > 0) cnt += n - 1;
	}
	cnt>>=1;
	return cnt<m;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);

	while (cin >> n >> m)
	{
		for (int i = 1; i <= n; i++) cin >> a[i];
		sort(a + 1, a + 1 + n);
		LL r=1e18, l=-1e18;
		LL ans;
		while (l <= r)
		{
			LL mid = l + r >> 1;
			if (check(mid)) {
				ans = mid;
				l = mid + 1;
			}
			else
				r = mid - 1;
		}
		cout << ans << endl;
	}

	return 0;
}
### 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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 2e5 + 2;
long long a[N];
long long n, m;
typedef long long LL;
bool check(LL x)
{
	long long h1, h2, cnt = 0;
	for (int i = 1; i <= n; i++)
	{
		if (a[i])
		{
			h1 = h2 = x / a[i];
			if (x % a[i])
			{
				if ((x < 0 && a[i]>0) || (x > 0 && a[i] < 0)) --h1;//不能直接相乘判断正负，因为会 爆
				else ++h2;
			}
			if (a[i] < 0)  cnt += n + 1 - (upper_bound(a + 1, a + 1 + n, h1) - a);
			if (a[i] > 0) cnt += lower_bound(a + 1, a + 1 + n, h2) - a - 1;
			if (a[i] * a[i] < x) cnt--;
		}
		else if (x > 0) cnt += n - 1;
	}
	cnt>>=1;
	return cnt<m;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);

	while (cin >> n >> m)
	{
		for (int i = 1; i <= n; i++) cin >> a[i];
		sort(a + 1, a + 1 + n);
		LL r=1e18, l=-1e18;
		LL ans;
		while (l <= r)
		{
			LL mid = l + r >> 1;
			if (check(mid)) {
				ans = mid;
				l = mid + 1;
			}
			else
				r = mid - 1;
		}
		cout << ans << endl;
	}

	return 0;
}
```