#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
ll a[maxn],n,k;
bool check(ll mid){//比它小的数大于等于k-1
    ll ans=0,x;
    for(int i=1;i<n;i++){
        if(a[i]<0){
            if(mid>0) x=mid/a[i];
            else x=mid/a[i]+(mid%a[i]!=0);
            ans+=n-(lower_bound(a+i+1,a+1+n,x)-(a+1));
        }
        else if(a[i]>0){
            if(mid<0) x=mid/a[i]-(mid%a[i]!=0);
            else x=mid/a[i];
            // ll x=floor(double(mid)/a[i]);
            ans+=upper_bound(a+i+1,a+1+n,x)-(a+i+1);
        }
        else{
            if(mid>=0) ans+=(n-i);
        }
    }
    return ans>=k;
}
int main(){
    cin>>n>>k;
    for(int i=1;i<=n;i++) cin>>a[i];
    sort(a+1,a+1+n);
    ll l=-1e18-3,r=1e18+3;
    while(l+1<r){
        ll mid=(l+r)>>1;
        if(check(mid)) r=mid;
        else l=mid;
    }
    cout<<r<<endl;
    return 0;
}
### Prompt

Your challenge is to write 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 int maxn=2e5+5;
ll a[maxn],n,k;
bool check(ll mid){//比它小的数大于等于k-1
    ll ans=0,x;
    for(int i=1;i<n;i++){
        if(a[i]<0){
            if(mid>0) x=mid/a[i];
            else x=mid/a[i]+(mid%a[i]!=0);
            ans+=n-(lower_bound(a+i+1,a+1+n,x)-(a+1));
        }
        else if(a[i]>0){
            if(mid<0) x=mid/a[i]-(mid%a[i]!=0);
            else x=mid/a[i];
            // ll x=floor(double(mid)/a[i]);
            ans+=upper_bound(a+i+1,a+1+n,x)-(a+i+1);
        }
        else{
            if(mid>=0) ans+=(n-i);
        }
    }
    return ans>=k;
}
int main(){
    cin>>n>>k;
    for(int i=1;i<=n;i++) cin>>a[i];
    sort(a+1,a+1+n);
    ll l=-1e18-3,r=1e18+3;
    while(l+1<r){
        ll mid=(l+r)>>1;
        if(check(mid)) r=mid;
        else l=mid;
    }
    cout<<r<<endl;
    return 0;
}
```