#include <bits/stdc++.h>
using namespace std;
#define rep(i, n)        for(int i=0; i<(int)(n); ++i)
#define ll               long long
//動画解説写し
const long long INF=1e18+1;
int main() {
  int n;
  ll k;
  cin>>n>>k;
  vector<ll>a(n);
  rep(i,n)cin>>a[i];
  sort(a.begin(),a.end());
  ll l=-INF,r=INF;
  while(l+1<r){
    ll x=(l+r)/2;
    bool ok=[&]{
      ll tot=0;
      rep(i,n){
        if(a[i]<0){
          int l=-1,r=n;
          while(l+1<r){
            int c=(l+r)/2;
            if(a[c]*a[i]<x)r=c;
            else l=c;
          }
          tot+=n-r;
        }else{
          int l=-1,r=n;
          while(l+1<r){
            int c=(l+r)/2;
            if(a[c]*a[i]<x)l=c;
            else r=c;
          }
          tot+=r;
        }
        if(a[i]*a[i]<x)tot--;
      }
      tot/=2;
      return tot<k;
    }();
    if(ok)l=x;
    else r=x;
  }
  cout<<l<<endl;
  return 0;
}
### 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<(int)(n); ++i)
#define ll               long long
//動画解説写し
const long long INF=1e18+1;
int main() {
  int n;
  ll k;
  cin>>n>>k;
  vector<ll>a(n);
  rep(i,n)cin>>a[i];
  sort(a.begin(),a.end());
  ll l=-INF,r=INF;
  while(l+1<r){
    ll x=(l+r)/2;
    bool ok=[&]{
      ll tot=0;
      rep(i,n){
        if(a[i]<0){
          int l=-1,r=n;
          while(l+1<r){
            int c=(l+r)/2;
            if(a[c]*a[i]<x)r=c;
            else l=c;
          }
          tot+=n-r;
        }else{
          int l=-1,r=n;
          while(l+1<r){
            int c=(l+r)/2;
            if(a[c]*a[i]<x)l=c;
            else r=c;
          }
          tot+=r;
        }
        if(a[i]*a[i]<x)tot--;
      }
      tot/=2;
      return tot<k;
    }();
    if(ok)l=x;
    else r=x;
  }
  cout<<l<<endl;
  return 0;
}
```