#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int N;
ll K,A[200000],L,R;
int main() 
{
	cin>>N>>K;
	for(int i=0;i<N;i++) 
	{
		cin>>A[i];
	}
	sort(A,A+N);
	L=-1e18;
	R=1e18;
	while(L+1!=R) 
	{
		ll M=(L+R)/2;
		ll cnt=0;
		for(int i=0;i<N;i++) 
		{
			if(A[i]<0) 
			{
				int l,r;
				l=i;
				r=N;
				while(l+1!=r) 
				{
					int m=(l+r)/2;
					if(A[i]*A[m]<=M) 
					{
						r=m;
					}
					else 
					{
						l=m;
					}
				}
				cnt+=N-r;
			}
			else 
			{
				int l,r;
				l=i;
				r=N;
				while(l+1!=r) 
				{
					int m=(l+r)/2;
					if(A[i]*A[m]<=M) 
					{
						l=m;
					}
					else 
					{
						r=m;
					}
				}
				cnt+=l-i;
			}
		}
		if(cnt<K)
		L=M;
		else 
		R=M;
	}
	cout<<R<<endl;
}


### 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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int N;
ll K,A[200000],L,R;
int main() 
{
	cin>>N>>K;
	for(int i=0;i<N;i++) 
	{
		cin>>A[i];
	}
	sort(A,A+N);
	L=-1e18;
	R=1e18;
	while(L+1!=R) 
	{
		ll M=(L+R)/2;
		ll cnt=0;
		for(int i=0;i<N;i++) 
		{
			if(A[i]<0) 
			{
				int l,r;
				l=i;
				r=N;
				while(l+1!=r) 
				{
					int m=(l+r)/2;
					if(A[i]*A[m]<=M) 
					{
						r=m;
					}
					else 
					{
						l=m;
					}
				}
				cnt+=N-r;
			}
			else 
			{
				int l,r;
				l=i;
				r=N;
				while(l+1!=r) 
				{
					int m=(l+r)/2;
					if(A[i]*A[m]<=M) 
					{
						l=m;
					}
					else 
					{
						r=m;
					}
				}
				cnt+=l-i;
			}
		}
		if(cnt<K)
		L=M;
		else 
		R=M;
	}
	cout<<R<<endl;
}


```