#include <bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll INF=ll(1e18)+1;
int main() {
  ios::sync_with_stdio(false);
	cin.tie(0);
  ll n,k;
  cin>>n>>k;
  vector<ll>a(n);
rep(i,n){
  cin>>a[i];
}
sort(a.begin(),a.end());
ll l=-INF;
ll r=INF;
while (l+1<r){
  ll cnt=0;
  ll x=(l+r)/2;
//cout<<x<<endl;


  rep(i,n){
//  ll i=1;
  //cout<<a[i];
    if (a[i]>=0){
      ll l1=-1; ll r1=n;
      while(l1+1<r1){
        ll c1=(l1+r1)/2;
        if(a[c1]*a[i]<x)l1=c1;else r1=c1;
      }
      cnt+=r1;
    }
    else {
      ll l1=-1; ll r1=n;
      while(l1+1<r1){
        ll c1=(l1+r1)/2;
        if(a[c1]*a[i]<x)r1=c1;else l1=c1;
      }
      cnt+=n-r1;
    //  cout<<cnt<<endl;
    }
    if(a[i]*a[i]<x)cnt-=1;

  }

  //if (ok)l=x;else r=x;
cnt/=2;
//cout<<cnt<<endl;
  if (cnt<k)l=x;else r=x;

}


cout<<l;
}

### Prompt

Your challenge is to write 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>
#define rep(i,n) for (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll INF=ll(1e18)+1;
int main() {
  ios::sync_with_stdio(false);
	cin.tie(0);
  ll n,k;
  cin>>n>>k;
  vector<ll>a(n);
rep(i,n){
  cin>>a[i];
}
sort(a.begin(),a.end());
ll l=-INF;
ll r=INF;
while (l+1<r){
  ll cnt=0;
  ll x=(l+r)/2;
//cout<<x<<endl;


  rep(i,n){
//  ll i=1;
  //cout<<a[i];
    if (a[i]>=0){
      ll l1=-1; ll r1=n;
      while(l1+1<r1){
        ll c1=(l1+r1)/2;
        if(a[c1]*a[i]<x)l1=c1;else r1=c1;
      }
      cnt+=r1;
    }
    else {
      ll l1=-1; ll r1=n;
      while(l1+1<r1){
        ll c1=(l1+r1)/2;
        if(a[c1]*a[i]<x)r1=c1;else l1=c1;
      }
      cnt+=n-r1;
    //  cout<<cnt<<endl;
    }
    if(a[i]*a[i]<x)cnt-=1;

  }

  //if (ok)l=x;else r=x;
cnt/=2;
//cout<<cnt<<endl;
  if (cnt<k)l=x;else r=x;

}


cout<<l;
}

```