#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int N;
long K;
long A[2<<17];
int main()
{
	cin>>N>>K;
	for(int i=0;i<N;i++)cin>>A[i];
	sort(A,A+N);
	long L=-2e18,R=2e18;
	while(R-L>1)
	{
		long M=(L+R)/2;
		long now=2*K;
		for(int i=0;i<N;i++)
		{
			if(A[i]*A[i]<=M)now++;
			if(A[i]==0)
			{
				if(M>=0)now-=N;
			}
			else if(A[i]>0)
			{
				if(M>=0)now-=upper_bound(A,A+N,M/A[i])-A;
				else
				{
					now-=upper_bound(A,A+N,-((-M+A[i]-1)/A[i]))-A;
				}
			}
			else
			{
				if(M>=0)now-=A+N-lower_bound(A,A+N,-(M/-A[i]));
				else now-=A+N-lower_bound(A,A+N,(-M-A[i]-1)/-A[i]);
			}
		}
		if(now<=0)R=M;
		else L=M;
	}
	cout<<R<<endl;
}

### Prompt

Generate 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
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int N;
long K;
long A[2<<17];
int main()
{
	cin>>N>>K;
	for(int i=0;i<N;i++)cin>>A[i];
	sort(A,A+N);
	long L=-2e18,R=2e18;
	while(R-L>1)
	{
		long M=(L+R)/2;
		long now=2*K;
		for(int i=0;i<N;i++)
		{
			if(A[i]*A[i]<=M)now++;
			if(A[i]==0)
			{
				if(M>=0)now-=N;
			}
			else if(A[i]>0)
			{
				if(M>=0)now-=upper_bound(A,A+N,M/A[i])-A;
				else
				{
					now-=upper_bound(A,A+N,-((-M+A[i]-1)/A[i]))-A;
				}
			}
			else
			{
				if(M>=0)now-=A+N-lower_bound(A,A+N,-(M/-A[i]));
				else now-=A+N-lower_bound(A,A+N,(-M-A[i]-1)/-A[i]);
			}
		}
		if(now<=0)R=M;
		else L=M;
	}
	cout<<R<<endl;
}

```