#include<bits/stdc++.h>
using namespace std;
#define pi 3.14159265359
#define int long long
#define pii pair<int,int>
const int mod=1e9+7;

int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1};




int32_t  main()
{ 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);


int n,k;
cin>>n>>k;
vector<int>v(n);

for(int i=0;i<n;i++)
{
  cin>>v[i];
}

sort(v.begin(),v.end());

    int l = -2e18;
    int r = 2e18;
    int ans=-1e18;
    while(r-l>1)
    {
        int M = (l+r)/2;
        int cnt = 0;
        
        for(int i = 0; i<n; ++i)
        {
            int l = i, r = n; 
            while(r-l>1)
            {
                int m = (l+r)/2;
                if(v[i]*v[m]<=M)
                    (v[i]>0) ? l=m : r = m;
                else
                    (v[i]>0) ? r = m : l = m;
                
            }
            if(v[i]>0) cnt+=l-i;
            else cnt+=n-r;
        }
        (cnt<k)?l= M:r=M;
        
    }
    cout<<r<<"\n";
  
}


### Prompt

Your task is to create 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 pi 3.14159265359
#define int long long
#define pii pair<int,int>
const int mod=1e9+7;

int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1};




int32_t  main()
{ 
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);


int n,k;
cin>>n>>k;
vector<int>v(n);

for(int i=0;i<n;i++)
{
  cin>>v[i];
}

sort(v.begin(),v.end());

    int l = -2e18;
    int r = 2e18;
    int ans=-1e18;
    while(r-l>1)
    {
        int M = (l+r)/2;
        int cnt = 0;
        
        for(int i = 0; i<n; ++i)
        {
            int l = i, r = n; 
            while(r-l>1)
            {
                int m = (l+r)/2;
                if(v[i]*v[m]<=M)
                    (v[i]>0) ? l=m : r = m;
                else
                    (v[i]>0) ? r = m : l = m;
                
            }
            if(v[i]>0) cnt+=l-i;
            else cnt+=n-r;
        }
        (cnt<k)?l= M:r=M;
        
    }
    cout<<r<<"\n";
  
}


```