#include<bits/stdc++.h>
using namespace std;
 
using lint=long long;
using Real=long double;
const lint INF=1e18;

int main()
{
   cin.tie(0);
   ios::sync_with_stdio(false);

   int N; lint K; cin>>N>>K;
   vector<int> a(N);
   for(int i=0;i<N;i++) cin>>a[i];
   sort(a.begin(),a.end());

   auto check=[&](lint x)
   {
      lint cnt=0;
      for(int i=0;i<N;i++){
         if(a[i]>0){
            lint mx=floor((Real)x/a[i]);
            cnt+=upper_bound(a.begin(),a.begin()+i,mx)-a.begin();
         }else if(a[i]<0){
            lint mn=ceil((Real)x/a[i]);
            cnt+=(a.begin()+i)-lower_bound(a.begin(),a.begin()+i,mn);
         }else{
            cnt+=(x>=0? i:0);
         }
      }
      return cnt>=K;
   };

   lint ng=-INF-1,ok=INF;
   while(ok-ng>1){
      lint mid=(ng+ok)/2;
      (check(mid)? ok:ng)=mid;
   }
   cout<<ok<<endl;
   return 0;
}

### 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;
 
using lint=long long;
using Real=long double;
const lint INF=1e18;

int main()
{
   cin.tie(0);
   ios::sync_with_stdio(false);

   int N; lint K; cin>>N>>K;
   vector<int> a(N);
   for(int i=0;i<N;i++) cin>>a[i];
   sort(a.begin(),a.end());

   auto check=[&](lint x)
   {
      lint cnt=0;
      for(int i=0;i<N;i++){
         if(a[i]>0){
            lint mx=floor((Real)x/a[i]);
            cnt+=upper_bound(a.begin(),a.begin()+i,mx)-a.begin();
         }else if(a[i]<0){
            lint mn=ceil((Real)x/a[i]);
            cnt+=(a.begin()+i)-lower_bound(a.begin(),a.begin()+i,mn);
         }else{
            cnt+=(x>=0? i:0);
         }
      }
      return cnt>=K;
   };

   lint ng=-INF-1,ok=INF;
   while(ok-ng>1){
      lint mid=(ng+ok)/2;
      (check(mid)? ok:ng)=mid;
   }
   cout<<ok<<endl;
   return 0;
}

```