#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
int32_t main()
{	
	#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	freopen("out","w",stdout);
	#endif

	int n,k;
	cin>>n>>k;
	vector<int> a(n);
	for(int &i:a)cin>>i;
	sort(a.begin(),a.end());
	int lo=-1e18,hi=1e18;
	while((hi-lo)>1)
	{
		int mid=(lo+hi)/2;
		int sum=0;
		for(int i=0;i<n;i++)
		{
			if(a[i]>0)
				sum+=lower_bound(a.begin(),a.end(),(long double)mid/a[i])-a.begin();
			if(a[i]<0)sum+=n-(upper_bound(a.begin(),a.end(),(long double)mid/a[i])-a.begin());
			if(a[i]==0)
			{
				if(mid>0)
				sum+=n;
			}
			if(a[i]*a[i]<mid)sum--;
		}
		if(sum/2<k){lo=mid;}
		else hi=mid;

	}
	cout<<lo<<"\n";
}

### 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;
typedef long long ll;
#define int ll
int32_t main()
{	
	#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	freopen("out","w",stdout);
	#endif

	int n,k;
	cin>>n>>k;
	vector<int> a(n);
	for(int &i:a)cin>>i;
	sort(a.begin(),a.end());
	int lo=-1e18,hi=1e18;
	while((hi-lo)>1)
	{
		int mid=(lo+hi)/2;
		int sum=0;
		for(int i=0;i<n;i++)
		{
			if(a[i]>0)
				sum+=lower_bound(a.begin(),a.end(),(long double)mid/a[i])-a.begin();
			if(a[i]<0)sum+=n-(upper_bound(a.begin(),a.end(),(long double)mid/a[i])-a.begin());
			if(a[i]==0)
			{
				if(mid>0)
				sum+=n;
			}
			if(a[i]*a[i]<mid)sum--;
		}
		if(sum/2<k){lo=mid;}
		else hi=mid;

	}
	cout<<lo<<"\n";
}

```