#include<bits/stdc++.h>
using namespace std;
long long a[200005],k;
int n;
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;++i)
	 cin>>a[i];
	sort(a,a+n);
	long long l=-1e18,r=1e18,m,res;
	while(l<=r)
	{
		m=(l+r)>>1;
		long long tot=0;
		for(int i=0;i<n;++i)
		{
			int lt=0,rt=n-1,mid;
			if(a[i]<0)
			{
				int ans=n;
				while(lt<=rt)
				{
					mid=(lt+rt)>>1;
					if(a[mid]*a[i]<=m) ans=mid,rt=mid-1;
					else lt=mid+1;
				}
				tot+=n-ans;
			}
			else
			{
				int ans=-1;
				while(lt<=rt)
				{
					mid=(lt+rt)>>1;
					if(a[mid]*a[i]<=m) ans=mid,lt=mid+1;
					else rt=mid-1;
				}
				tot+=ans+1;
			}
			if(a[i]*a[i]<=m) --tot;
		}
		if((tot/2)>=k) res=m,r=m-1;
		else l=m+1;
	}
	cout<<res;
	return 0;
}
### Prompt

Develop a solution in Cpp 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<bits/stdc++.h>
using namespace std;
long long a[200005],k;
int n;
int main()
{
	cin>>n>>k;
	for(int i=0;i<n;++i)
	 cin>>a[i];
	sort(a,a+n);
	long long l=-1e18,r=1e18,m,res;
	while(l<=r)
	{
		m=(l+r)>>1;
		long long tot=0;
		for(int i=0;i<n;++i)
		{
			int lt=0,rt=n-1,mid;
			if(a[i]<0)
			{
				int ans=n;
				while(lt<=rt)
				{
					mid=(lt+rt)>>1;
					if(a[mid]*a[i]<=m) ans=mid,rt=mid-1;
					else lt=mid+1;
				}
				tot+=n-ans;
			}
			else
			{
				int ans=-1;
				while(lt<=rt)
				{
					mid=(lt+rt)>>1;
					if(a[mid]*a[i]<=m) ans=mid,lt=mid+1;
					else rt=mid-1;
				}
				tot+=ans+1;
			}
			if(a[i]*a[i]<=m) --tot;
		}
		if((tot/2)>=k) res=m,r=m-1;
		else l=m+1;
	}
	cout<<res;
	return 0;
}
```