#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 1000000007
#define INF 1e9
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()

ll intceil(ll a,ll b){
    if(a>=0) return (a+b-1)/b;
    else return a/b;
}

ll intfloor(ll a,ll b){
    if(a<0) return (a-b+1)/b;
    else return a/b;
}

int main(){
    ll N,K;
    cin>>N>>K;
    vector<ll> A(N);
    rep(i,N) cin>>A[i];
    sort(all(A));

    ll ok=1e18,ng=-1e18-1;
    while(ok-ng>1){
        ll m=(ok+ng)/2;
        ll num=0;
        rep(i,N){
            if(A[i]<0){
                ll v=intceil(-m,-A[i]);
                num+=A.begin()+i-lower_bound(A.begin(),A.begin()+i,v);
            }
            else if(A[i]==0){
                if(m>=0) num+=i;
            }
            else{
                ll v=intfloor(m,A[i]);
                num+=upper_bound(A.begin(),A.begin()+i,v)-A.begin();
            }
        }
        (num>=K ? ok : ng)=m;
    }

    cout<<ok<<endl;
}
### Prompt

In CPP, your task is to solve 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;
#define MOD 1000000007
#define INF 1e9
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()

ll intceil(ll a,ll b){
    if(a>=0) return (a+b-1)/b;
    else return a/b;
}

ll intfloor(ll a,ll b){
    if(a<0) return (a-b+1)/b;
    else return a/b;
}

int main(){
    ll N,K;
    cin>>N>>K;
    vector<ll> A(N);
    rep(i,N) cin>>A[i];
    sort(all(A));

    ll ok=1e18,ng=-1e18-1;
    while(ok-ng>1){
        ll m=(ok+ng)/2;
        ll num=0;
        rep(i,N){
            if(A[i]<0){
                ll v=intceil(-m,-A[i]);
                num+=A.begin()+i-lower_bound(A.begin(),A.begin()+i,v);
            }
            else if(A[i]==0){
                if(m>=0) num+=i;
            }
            else{
                ll v=intfloor(m,A[i]);
                num+=upper_bound(A.begin(),A.begin()+i,v)-A.begin();
            }
        }
        (num>=K ? ok : ng)=m;
    }

    cout<<ok<<endl;
}
```