#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
using ll =long long;

const ll INF =ll(1e18)+1;

int main(){
  ll n,k;
  cin >> n >>k;
  vector<ll> a(n);
  rep(i,n)cin >> a.at(i);
  sort(a.begin(),a.end());
  ll l = -INF,r=INF;
  while(l+1<r){  
  	ll x = (l+r)/2,count=0; 
    rep(i,n){
      if(a[i]<0){
        int l=-1,r=n;
        while (l+1<r) {
          int m = (l+r)/2;
          if (a[i] * a[m] < x) {
            r = m;
          } else {
            l = m;
          }
        }
        count+=n-r;
      }else{
        int l=-1,r=n;
        while(l+1<r){
          int m =(l+r)/2;
          if(a[i]*a[m]<x){
            l = m;
          }else{
            r = m;
          }
        }       
        count+=r;   
      }
      if(a[i]*a[i]<x)count--;
    }
    count/=2;
    if(count<k){
      l=x;
    }else{
      r=x;
    }
  }
  cout << l << endl;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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++)
using ll =long long;

const ll INF =ll(1e18)+1;

int main(){
  ll n,k;
  cin >> n >>k;
  vector<ll> a(n);
  rep(i,n)cin >> a.at(i);
  sort(a.begin(),a.end());
  ll l = -INF,r=INF;
  while(l+1<r){  
  	ll x = (l+r)/2,count=0; 
    rep(i,n){
      if(a[i]<0){
        int l=-1,r=n;
        while (l+1<r) {
          int m = (l+r)/2;
          if (a[i] * a[m] < x) {
            r = m;
          } else {
            l = m;
          }
        }
        count+=n-r;
      }else{
        int l=-1,r=n;
        while(l+1<r){
          int m =(l+r)/2;
          if(a[i]*a[m]<x){
            l = m;
          }else{
            r = m;
          }
        }       
        count+=r;   
      }
      if(a[i]*a[i]<x)count--;
    }
    count/=2;
    if(count<k){
      l=x;
    }else{
      r=x;
    }
  }
  cout << l << endl;
}
```