#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i=0;i<(n);++i)
#define fi first
#define se second
#define INF 1000000009
#define lINF 1000000000000000009
typedef long long ll;
ll n,k;
ll a[200009];

bool judge(ll mid){
  ll cnt=0;
  rep(i,n){
    if(a[i]<0){
      int l=-1,r=n;
      while(r-l>1){
        int j=(l+r)/2;
        if(a[i]*a[j]<mid){
          r=j;
        }
        else{
          l=j;
        }
      }
      cnt+=n-r;
    }
    else{
      int l=-1,r=n;
      while(r-l>1){
        ll j=(l+r)/2;
        if(a[i]*a[j]<mid){
          l=j;
        }
        else{
          r=j;
        }
      }
      cnt+=r;
    }
    if(a[i]*a[i]<mid)cnt--;
  }
  cnt/=2;

  return cnt<k;
}


int main(){
  cin>>n>>k;
  rep(i,n){
    cin>>a[i];
  }
  sort(a,a+n);

  ll l=-lINF,r=lINF;
  while(r-l>1){
    ll mid=(l+r)/2;
    if(judge(mid)){
      l=mid;
    }
    else{
      r=mid;
    }
  }

  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<(n);++i)
#define fi first
#define se second
#define INF 1000000009
#define lINF 1000000000000000009
typedef long long ll;
ll n,k;
ll a[200009];

bool judge(ll mid){
  ll cnt=0;
  rep(i,n){
    if(a[i]<0){
      int l=-1,r=n;
      while(r-l>1){
        int j=(l+r)/2;
        if(a[i]*a[j]<mid){
          r=j;
        }
        else{
          l=j;
        }
      }
      cnt+=n-r;
    }
    else{
      int l=-1,r=n;
      while(r-l>1){
        ll j=(l+r)/2;
        if(a[i]*a[j]<mid){
          l=j;
        }
        else{
          r=j;
        }
      }
      cnt+=r;
    }
    if(a[i]*a[i]<mid)cnt--;
  }
  cnt/=2;

  return cnt<k;
}


int main(){
  cin>>n>>k;
  rep(i,n){
    cin>>a[i];
  }
  sort(a,a+n);

  ll l=-lINF,r=lINF;
  while(r-l>1){
    ll mid=(l+r)/2;
    if(judge(mid)){
      l=mid;
    }
    else{
      r=mid;
    }
  }

  cout<<l<<endl;

  return(0);
}

```