#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

int binary_search(int n,ll *a,int i,ll trg){
  int l,r,mid;
  l=-1;
  r=n;
  
  if (a[i]>=0){
    while (r-l>1){
      mid=(l+r)/2;
      if (a[i]*a[mid] <= trg) l=mid;
      else r=mid;
    }
    if (l>=i) return l;
    else return l+1;
  }else{
    while (r-l>1){
      mid=(l+r+1)/2;
      if (a[i]*a[mid] <= trg) r=mid;
      else l=mid;
    }
    if (r<=i) return n-r-1;
    else return n-r;
  }
}

int main(void){
  int n,i;
  ll a[200010],k;
  cin >> n >> k;
  for (i=0;i<n;i++) cin >> a[i];
  sort(a,a+n);
  
  ll l,r,mid,s;
  l=-1e18;
  r=1e18;
  while (r-l>1){
    mid=(l+r)/2;
    s=0;
    for (i=0;i<n;i++){
      s+=binary_search(n,a,i,mid);
    }
    s/=2;
    if (s>=k) r=mid;
    else l=mid;
  }
  
  cout << r << endl;
  
  return 0;
}
### Prompt

Please create a solution in cpp 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 <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

int binary_search(int n,ll *a,int i,ll trg){
  int l,r,mid;
  l=-1;
  r=n;
  
  if (a[i]>=0){
    while (r-l>1){
      mid=(l+r)/2;
      if (a[i]*a[mid] <= trg) l=mid;
      else r=mid;
    }
    if (l>=i) return l;
    else return l+1;
  }else{
    while (r-l>1){
      mid=(l+r+1)/2;
      if (a[i]*a[mid] <= trg) r=mid;
      else l=mid;
    }
    if (r<=i) return n-r-1;
    else return n-r;
  }
}

int main(void){
  int n,i;
  ll a[200010],k;
  cin >> n >> k;
  for (i=0;i<n;i++) cin >> a[i];
  sort(a,a+n);
  
  ll l,r,mid,s;
  l=-1e18;
  r=1e18;
  while (r-l>1){
    mid=(l+r)/2;
    s=0;
    for (i=0;i<n;i++){
      s+=binary_search(n,a,i,mid);
    }
    s/=2;
    if (s>=k) r=mid;
    else l=mid;
  }
  
  cout << r << endl;
  
  return 0;
}
```