#include <bits/stdc++.h>
using namespace std;

using ll = long long;


ll N,K;

int my_b_search(ll key,ll val,bool mode,vector<int64_t>& A){
    //mode -> true ならば昇順での二分探索 falseなら降順での二分探索
    int ok = mode ? -1 : N;
    int ng = mode ? N : -1;
    while(abs(ok-ng)>1){
        int mid=(ok+ng)/2;
        if(val*A[mid]<=key)ok=mid;
        else ng=mid;
    }
    return ok;
}

bool f(ll key,vector<int64_t>& A){
    ll cnt=0;
    for(int i=0;i<N;i++){
        if(A[i]>=0){
            cnt+=my_b_search(key,A[i],true,A)+1;
        }else{
            cnt+=N-my_b_search(key,A[i],false,A);
        }
        if(key>=A[i]*A[i])cnt--;
    }
    return cnt>=2*K;
}


int main(){
    
    cin>>N>>K;
    vector<int64_t> A(N);

    for(int64_t& a:A){
        cin>>a;
    }

    sort(A.begin(),A.end());

    int64_t ok=1e18,ng=-1e18;
    while(ok-ng!=1){
        int64_t mid=(ok+ng)/2;

        if(f(mid,A)){
            ok=mid;
        }else{
            ng=mid;
        }
    }

    cout<<ok<<endl;

}
### Prompt

Create a solution in CPP for 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 ll = long long;


ll N,K;

int my_b_search(ll key,ll val,bool mode,vector<int64_t>& A){
    //mode -> true ならば昇順での二分探索 falseなら降順での二分探索
    int ok = mode ? -1 : N;
    int ng = mode ? N : -1;
    while(abs(ok-ng)>1){
        int mid=(ok+ng)/2;
        if(val*A[mid]<=key)ok=mid;
        else ng=mid;
    }
    return ok;
}

bool f(ll key,vector<int64_t>& A){
    ll cnt=0;
    for(int i=0;i<N;i++){
        if(A[i]>=0){
            cnt+=my_b_search(key,A[i],true,A)+1;
        }else{
            cnt+=N-my_b_search(key,A[i],false,A);
        }
        if(key>=A[i]*A[i])cnt--;
    }
    return cnt>=2*K;
}


int main(){
    
    cin>>N>>K;
    vector<int64_t> A(N);

    for(int64_t& a:A){
        cin>>a;
    }

    sort(A.begin(),A.end());

    int64_t ok=1e18,ng=-1e18;
    while(ok-ng!=1){
        int64_t mid=(ok+ng)/2;

        if(f(mid,A)){
            ok=mid;
        }else{
            ng=mid;
        }
    }

    cout<<ok<<endl;

}
```