#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[200005],n,k;
map<ll,ll> p;
bool check(ll x){
    ll num=0;
    for(ll i=1;i<n;i++){
        if(a[i]*a[i+1]>x&&a[i]*a[n]>x) continue;
        if(a[i]==0){
            num+=n-i;continue;
        }
        ll l=1,r=n-i;
        while(r>=l){
            ll mid=(l+r)/2;
            if(a[i]>0){
                if(a[i]*a[i+mid]>x) r=mid-1;
                else l=mid+1;
            }
            else{
                if(a[i]*a[n-mid+1]>x) r=mid-1;
                else l=mid+1;
            }
        }
        num+=r;
    }
    return num<k;
}
int main(){
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        p[a[i]]=1;
    }
    sort(a+1,a+1+n);
    ll l=-1e18,r=1e18;
    while(r>=l){
        ll mid=(l+r)/2;
        if(check(mid)) l=mid+1;
        else r=mid-1;
    }
    cout<<l<<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;
ll a[200005],n,k;
map<ll,ll> p;
bool check(ll x){
    ll num=0;
    for(ll i=1;i<n;i++){
        if(a[i]*a[i+1]>x&&a[i]*a[n]>x) continue;
        if(a[i]==0){
            num+=n-i;continue;
        }
        ll l=1,r=n-i;
        while(r>=l){
            ll mid=(l+r)/2;
            if(a[i]>0){
                if(a[i]*a[i+mid]>x) r=mid-1;
                else l=mid+1;
            }
            else{
                if(a[i]*a[n-mid+1]>x) r=mid-1;
                else l=mid+1;
            }
        }
        num+=r;
    }
    return num<k;
}
int main(){
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        p[a[i]]=1;
    }
    sort(a+1,a+1+n);
    ll l=-1e18,r=1e18;
    while(r>=l){
        ll mid=(l+r)/2;
        if(check(mid)) l=mid+1;
        else r=mid-1;
    }
    cout<<l<<endl;
}
```