#include<bits/stdc++.h>
using namespace std;
#define rep(i,N) for(int i=0;i<(long long)N;i++)
typedef int64_t LL;
int main()
{
  int NG,OK,BASE;
  LL N,K,L=-pow(16,16)/16,M,R=LLONG_MAX/3,count;
  bool minus;
  cin>>N>>K;
  vector<LL> A(N,0);
  rep(i,N)cin>>A.at(i);
  sort(A.begin(),A.end());
  while(R-L>1)
  {
    M=(L+R)/2,count=0;
    rep(i,N)
    {
      NG=i,OK=N;
      minus=(A.at(i)<0)?true:false;
      while(OK-NG>1)
      {
        BASE=(NG+OK)/2;
        if(A.at(i)*A.at(BASE)<=M)(minus)?OK=BASE:NG=BASE;
        else(minus)?NG=BASE:OK=BASE;
      }
      count+=(minus)?N-OK:NG-i;
    }
    (count<K)?L=M:R=M;
  }
  cout<<R<<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 rep(i,N) for(int i=0;i<(long long)N;i++)
typedef int64_t LL;
int main()
{
  int NG,OK,BASE;
  LL N,K,L=-pow(16,16)/16,M,R=LLONG_MAX/3,count;
  bool minus;
  cin>>N>>K;
  vector<LL> A(N,0);
  rep(i,N)cin>>A.at(i);
  sort(A.begin(),A.end());
  while(R-L>1)
  {
    M=(L+R)/2,count=0;
    rep(i,N)
    {
      NG=i,OK=N;
      minus=(A.at(i)<0)?true:false;
      while(OK-NG>1)
      {
        BASE=(NG+OK)/2;
        if(A.at(i)*A.at(BASE)<=M)(minus)?OK=BASE:NG=BASE;
        else(minus)?NG=BASE:OK=BASE;
      }
      count+=(minus)?N-OK:NG-i;
    }
    (count<K)?L=M:R=M;
  }
  cout<<R<<endl;
}
```