#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define mp make_pair
typedef pair<int,int> pii;
ll a[200009];
ll n;ll k;
int main()
{
    scanf("%d%lld",&n,&k);
    for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
    sort(a+1,a+n+1);
    ll l=-1e18,r=1e18;
    ll answer;
    while(l<=r)
    {
        ll mid=(l+r)/2;ll cnt=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]>0){
                cnt+=upper_bound(a+1,a+n+1,(long double)mid/a[i])-a-1;
            }
            else if(a[i]<0){
                cnt+=n-(lower_bound(a+1,a+n+1,(long double)mid/a[i])-a-1);
            }
            else if(mid>=0)cnt+=n;
            if(a[i]*a[i]<=mid)cnt--;
        }
        if(cnt>=k*2)answer=mid,r=mid-1;
        else l=mid+1;
    }
    cout<<answer<<endl;
    return 0;
}
### Prompt

Please formulate 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;
#define ll long long
#define ull unsigned long long
#define mp make_pair
typedef pair<int,int> pii;
ll a[200009];
ll n;ll k;
int main()
{
    scanf("%d%lld",&n,&k);
    for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
    sort(a+1,a+n+1);
    ll l=-1e18,r=1e18;
    ll answer;
    while(l<=r)
    {
        ll mid=(l+r)/2;ll cnt=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]>0){
                cnt+=upper_bound(a+1,a+n+1,(long double)mid/a[i])-a-1;
            }
            else if(a[i]<0){
                cnt+=n-(lower_bound(a+1,a+n+1,(long double)mid/a[i])-a-1);
            }
            else if(mid>=0)cnt+=n;
            if(a[i]*a[i]<=mid)cnt--;
        }
        if(cnt>=k*2)answer=mid,r=mid-1;
        else l=mid+1;
    }
    cout<<answer<<endl;
    return 0;
}
```