#include<bits/stdc++.h>
using namespace std;
#define fort(i,a,n) for(i=a;i<=n;i++)
#define ford(i,n,a) for(i=n;i>=a;i--)
typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned long long ull;
const int N=3e5;
ll n,m,i,l,r,mid,anss,a[N],bo;
ll pd(ll k)
{
	ll i,l,r,mid,ans;
	ans=0;
	fort(i,1,n)
	{
		if (a[i]<0)
		{
			l=i+1; r=n;
			bo=i;
			while (l<=r)
			{
				mid=(l+r)/2;
				if (a[mid]*a[i]>=k)
				{
					l=mid+1;
					bo=mid;
				} else r=mid-1;
			}
			ans+=n-bo;
		} else {
			l=i+1; r=n;
			bo=i;
			while (l<=r)
			{
				mid=(l+r)/2;
				if (a[mid]*a[i]<k)
				{
					l=mid+1;
					bo=mid;
			    }else {
			    	r=mid-1;	
			    }
			}
			ans+=bo-i;
		}
	}
	return ans;
}//how many small
int main()
{
	cin>>n>>m;
	fort(i,1,n) 
	  scanf("%lld",&a[i]);
	sort(a+1,a+1+n);
	l=-1e18; r=1e18;
	while (l<=r)
	{
		mid=(l+r)/2;
		if (pd(mid)<m)
		{
			l=mid+1;
			anss=mid;
		} else r=mid-1;
	}
	cout<<anss<<endl;
}
### Prompt

Please formulate 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;
#define fort(i,a,n) for(i=a;i<=n;i++)
#define ford(i,n,a) for(i=n;i>=a;i--)
typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned long long ull;
const int N=3e5;
ll n,m,i,l,r,mid,anss,a[N],bo;
ll pd(ll k)
{
	ll i,l,r,mid,ans;
	ans=0;
	fort(i,1,n)
	{
		if (a[i]<0)
		{
			l=i+1; r=n;
			bo=i;
			while (l<=r)
			{
				mid=(l+r)/2;
				if (a[mid]*a[i]>=k)
				{
					l=mid+1;
					bo=mid;
				} else r=mid-1;
			}
			ans+=n-bo;
		} else {
			l=i+1; r=n;
			bo=i;
			while (l<=r)
			{
				mid=(l+r)/2;
				if (a[mid]*a[i]<k)
				{
					l=mid+1;
					bo=mid;
			    }else {
			    	r=mid-1;	
			    }
			}
			ans+=bo-i;
		}
	}
	return ans;
}//how many small
int main()
{
	cin>>n>>m;
	fort(i,1,n) 
	  scanf("%lld",&a[i]);
	sort(a+1,a+1+n);
	l=-1e18; r=1e18;
	while (l<=r)
	{
		mid=(l+r)/2;
		if (pd(mid)<m)
		{
			l=mid+1;
			anss=mid;
		} else r=mid-1;
	}
	cout<<anss<<endl;
}
```