#include <bits/stdc++.h>
using namespace std;
const int N=1e6+10;
#define int long long
vector<int>a(N);
int32_t main() 
{
    int n,k; cin>>n>>k; a.resize(n);
    for(int i=0;i<n;i++)  cin>>a[i];
    sort(a.begin(),a.end());
    
    int l=-1e18+1,r=1e18+1;
    while(r-l>1)
    {
        int mid=(l+r)>>1;  int ans=0;
        for(int i=0;i<n;i++)
        {
          if(a[i]<0)
         {
            int lx=-1,rn=n;
            while(rn-lx>1)
            {
                int curr=(rn+lx)>>1;
                if(a[curr]*a[i]<mid) rn=curr;
                else lx=curr; 
            }
             ans+=(n-rn);
         }
        
        else
        {
            int lx=-1,rn=n;
            while(rn-lx>1)
            {
                int curr=(rn+lx)>>1;
                if(a[curr]*a[i]<mid) lx=curr;
                else rn=curr;
            }
          ans+=rn;
        }
         if(a[i]*a[i]<mid) ans--;
    }
       ans/=2;
       if(ans<k) l=mid;
       else r=mid; 
    }
    cout<<l; return 0;
}
### Prompt

Generate 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;
const int N=1e6+10;
#define int long long
vector<int>a(N);
int32_t main() 
{
    int n,k; cin>>n>>k; a.resize(n);
    for(int i=0;i<n;i++)  cin>>a[i];
    sort(a.begin(),a.end());
    
    int l=-1e18+1,r=1e18+1;
    while(r-l>1)
    {
        int mid=(l+r)>>1;  int ans=0;
        for(int i=0;i<n;i++)
        {
          if(a[i]<0)
         {
            int lx=-1,rn=n;
            while(rn-lx>1)
            {
                int curr=(rn+lx)>>1;
                if(a[curr]*a[i]<mid) rn=curr;
                else lx=curr; 
            }
             ans+=(n-rn);
         }
        
        else
        {
            int lx=-1,rn=n;
            while(rn-lx>1)
            {
                int curr=(rn+lx)>>1;
                if(a[curr]*a[i]<mid) lx=curr;
                else rn=curr;
            }
          ans+=rn;
        }
         if(a[i]*a[i]<mid) ans--;
    }
       ans/=2;
       if(ans<k) l=mid;
       else r=mid; 
    }
    cout<<l; return 0;
}
```