#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 1e18;
ll n,k;
bool calc(ll x,vector<ll> &vec){
  
  ll num=0;
  
  for(int i=0;i<n;i++){
    int left=i,right=n;
    
    if(vec[i]>=0) {
      while(left+1<right){
        int mid=(left+right)/2;
        if(vec[i]*vec[mid]<=x) left=mid;
        else right=mid;
      }
      num+=left-i;
    }
    
    else if(vec[i]<0){
      while(left+1<right){
        int mid=(left+right)/2;
        if(vec[i]*vec[mid]<=x) right=mid;
        else left=mid;
      }
      num+=n-right;
    }
    
  }
  
  return num<k;
}

int main(){
  cin >> n >> k;
  vector<ll> vec(n);
  for(int i=0;i<n;i++){
    cin >> vec[i];
  }
  sort(vec.begin(),vec.end());
  ll ok = INF+1,ng = -INF-1;
  while(ok>ng+1){
    ll mid = (ok+ng)/2;
    if(calc(mid,vec)) ng=mid;
    else ok=mid;
  }
  cout << ok << endl;
}
### Prompt

Your task is to create 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;
typedef long long ll;
const ll INF = 1e18;
ll n,k;
bool calc(ll x,vector<ll> &vec){
  
  ll num=0;
  
  for(int i=0;i<n;i++){
    int left=i,right=n;
    
    if(vec[i]>=0) {
      while(left+1<right){
        int mid=(left+right)/2;
        if(vec[i]*vec[mid]<=x) left=mid;
        else right=mid;
      }
      num+=left-i;
    }
    
    else if(vec[i]<0){
      while(left+1<right){
        int mid=(left+right)/2;
        if(vec[i]*vec[mid]<=x) right=mid;
        else left=mid;
      }
      num+=n-right;
    }
    
  }
  
  return num<k;
}

int main(){
  cin >> n >> k;
  vector<ll> vec(n);
  for(int i=0;i<n;i++){
    cin >> vec[i];
  }
  sort(vec.begin(),vec.end());
  ll ok = INF+1,ng = -INF-1;
  while(ok>ng+1){
    ll mid = (ok+ng)/2;
    if(calc(mid,vec)) ng=mid;
    else ok=mid;
  }
  cout << ok << endl;
}
```