#include<bits/stdc++.h>
using namespace std;
#define int long long
const long long inf=1e18+1;
int n,k;
int a[200001];
bool judge(long long x)
{
	long long ans=0;
	for(int i=1;i<=n;i++)
	{
		if(a[i]<0)
		{
			int tmp=x/a[i];
			if(x>0&&abs(x)%a[i]!=0)
				tmp--;
			int sum=n-max(i,(long long)(upper_bound(a+1,a+n+1,tmp-(abs(x)%abs(a[i])==0))-a-1));
			ans+=max(0ll,sum);
		}
		else if(a[i]>0)
		{
			int tmp=x/a[i];
			if(x<0&&abs(x)%a[i]!=0)
				tmp--;
			int sum=upper_bound(a+1,a+n+1,tmp)-a-1;
			ans+=max(0ll,sum-i);
		}
		else if(a[i]==0)
		{
			if(x>=0)
				ans+=n-i;
		}
	}
	return ans>=k;
}
signed main()
{
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	long long l=-inf-1,r=inf+1;
	while(l+1<r)
	{
		long long mid=(l+r)/2;
		if(judge(mid))
			r=mid;
		else
			l=mid;
	}
	printf("%lld",r);
	return 0;
}
/*
4 6
-1 0 1 2
*/
### Prompt

In cpp, your task is to solve 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;
#define int long long
const long long inf=1e18+1;
int n,k;
int a[200001];
bool judge(long long x)
{
	long long ans=0;
	for(int i=1;i<=n;i++)
	{
		if(a[i]<0)
		{
			int tmp=x/a[i];
			if(x>0&&abs(x)%a[i]!=0)
				tmp--;
			int sum=n-max(i,(long long)(upper_bound(a+1,a+n+1,tmp-(abs(x)%abs(a[i])==0))-a-1));
			ans+=max(0ll,sum);
		}
		else if(a[i]>0)
		{
			int tmp=x/a[i];
			if(x<0&&abs(x)%a[i]!=0)
				tmp--;
			int sum=upper_bound(a+1,a+n+1,tmp)-a-1;
			ans+=max(0ll,sum-i);
		}
		else if(a[i]==0)
		{
			if(x>=0)
				ans+=n-i;
		}
	}
	return ans>=k;
}
signed main()
{
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	long long l=-inf-1,r=inf+1;
	while(l+1<r)
	{
		long long mid=(l+r)/2;
		if(judge(mid))
			r=mid;
		else
			l=mid;
	}
	printf("%lld",r);
	return 0;
}
/*
4 6
-1 0 1 2
*/
```