#include<bits/stdc++.h>
using namespace std;

bool check(long long a[],long long mid,long long n,long long k)
{
  long long i,s=0;
  for(i=0;i<n;i++)
  {
    if(a[i]>0)
    {
      s+=lower_bound(a,a+n,(long double)mid/(a[i]))-a;
    }
    if(a[i]<0)
    {
      s+=n-(upper_bound(a,a+n,(long double)mid/(a[i]))-a);
    }
    if(a[i]==0)
    {
      s+=(mid>0)?n:0;
    }
    if(a[i]*a[i]<mid)
      s--;
  }
    if((s/2)<k)
      return true;
    else
      return false;
  
}

int main()
{

    ios_base::sync_with_stdio(false);
      cin.tie(NULL);cout.tie(NULL);
      long long n,k;
      cin>>n>>k;
      long long a[n],i;
      for(i=0;i<n;i++)
      {
        cin>>a[i];
      }
      sort(a,a+n);
      long long low=-2e18,high=2e18,mid;
      while(high-low>1)
      {
        long long mid=(low+high)/2;
        if(check(a,mid,n,k))
          low=mid;
        else
          high=mid;
      }
      if(check(a,low,n,k))
      cout<<low<<endl;
      else
        cout<<high<<endl;
}
### 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;

bool check(long long a[],long long mid,long long n,long long k)
{
  long long i,s=0;
  for(i=0;i<n;i++)
  {
    if(a[i]>0)
    {
      s+=lower_bound(a,a+n,(long double)mid/(a[i]))-a;
    }
    if(a[i]<0)
    {
      s+=n-(upper_bound(a,a+n,(long double)mid/(a[i]))-a);
    }
    if(a[i]==0)
    {
      s+=(mid>0)?n:0;
    }
    if(a[i]*a[i]<mid)
      s--;
  }
    if((s/2)<k)
      return true;
    else
      return false;
  
}

int main()
{

    ios_base::sync_with_stdio(false);
      cin.tie(NULL);cout.tie(NULL);
      long long n,k;
      cin>>n>>k;
      long long a[n],i;
      for(i=0;i<n;i++)
      {
        cin>>a[i];
      }
      sort(a,a+n);
      long long low=-2e18,high=2e18,mid;
      while(high-low>1)
      {
        long long mid=(low+high)/2;
        if(check(a,mid,n,k))
          low=mid;
        else
          high=mid;
      }
      if(check(a,low,n,k))
      cout<<low<<endl;
      else
        cout<<high<<endl;
}
```