#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=2e5+5,inf=1e18+5;
ll n,k,a[N];
bool chk(ll x){
    ll tot=0;
    for(ll i=0;i<n;i++){
        if(a[i]<0){
            ll l=-1,r=n;
            while(l+1<r){
                ll mid=(l+r)/2;
                if(a[mid]*a[i]<x)r=mid;
                else l=mid;
            }
            tot+=n-r;
        }
        else{
            ll l=-1,r=n;
            while(l+1<r){
                ll mid=(l+r)/2;
                if(a[mid]*a[i]<x)l=mid;
                else r=mid;
            }
            tot+=r;
        }
        if(a[i]*a[i]<x)tot--;
    }
    tot/=2;
    return tot<k;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n>>k;
    for(ll i=0;i<n;i++)cin>>a[i];
    sort(a,a+n);
    ll l=-inf,r=inf;
    while(l+1<r){
        ll mid=(l+r)/2;
        if(chk(mid))l=mid;
        else r=mid;
    }
    cout<<l;
}

### Prompt

Generate 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 N=2e5+5,inf=1e18+5;
ll n,k,a[N];
bool chk(ll x){
    ll tot=0;
    for(ll i=0;i<n;i++){
        if(a[i]<0){
            ll l=-1,r=n;
            while(l+1<r){
                ll mid=(l+r)/2;
                if(a[mid]*a[i]<x)r=mid;
                else l=mid;
            }
            tot+=n-r;
        }
        else{
            ll l=-1,r=n;
            while(l+1<r){
                ll mid=(l+r)/2;
                if(a[mid]*a[i]<x)l=mid;
                else r=mid;
            }
            tot+=r;
        }
        if(a[i]*a[i]<x)tot--;
    }
    tot/=2;
    return tot<k;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n>>k;
    for(ll i=0;i<n;i++)cin>>a[i];
    sort(a,a+n);
    ll l=-inf,r=inf;
    while(l+1<r){
        ll mid=(l+r)/2;
        if(chk(mid))l=mid;
        else r=mid;
    }
    cout<<l;
}

```