#include<bits/stdc++.h>
using namespace std;
int main() {
  long long n,k,i,l,r,m,c,ll,rr,mm;
  cin>>n>>k;
  vector<long long> a(n);
  for (i=0;i<n;i++) cin>>a.at(i);
  sort(a.begin(),a.end());
  l=-1e18;
  r=1e18;
  while (r-l>1) {
    m=(l+r)/2;
    c=0;
    for (i=0;i<n;i++) {
      if (a.at(i)<0) {
        if (a.at(i)*a.at(n-1)>m) continue;
        ll=-1;rr=n-1;
        while (rr-ll>1) {
          mm=(ll+rr)/2;
          if (a.at(i)*a.at(mm)<=m) rr=mm;
          else ll=mm;
        }
        c+=n-rr;
      }
      else if (a.at(i)==0 && 0<=m) c+=n;
      else if (a.at(i)>0) {
        if (a.at(i)*a.at(0)>m) continue;
        ll=0;rr=n;
        while (rr-ll>1) {
          mm=(ll+rr)/2;
          if (a.at(i)*a.at(mm)<=m) ll=mm;
          else rr=mm;
        }
        c+=ll+1;
      }
      if (a.at(i)*a.at(i)<=m) c--;
    }
    if (c<2*k) l=m;
    else r=m;
  }
  cout<<r<<endl;
}
### 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;
int main() {
  long long n,k,i,l,r,m,c,ll,rr,mm;
  cin>>n>>k;
  vector<long long> a(n);
  for (i=0;i<n;i++) cin>>a.at(i);
  sort(a.begin(),a.end());
  l=-1e18;
  r=1e18;
  while (r-l>1) {
    m=(l+r)/2;
    c=0;
    for (i=0;i<n;i++) {
      if (a.at(i)<0) {
        if (a.at(i)*a.at(n-1)>m) continue;
        ll=-1;rr=n-1;
        while (rr-ll>1) {
          mm=(ll+rr)/2;
          if (a.at(i)*a.at(mm)<=m) rr=mm;
          else ll=mm;
        }
        c+=n-rr;
      }
      else if (a.at(i)==0 && 0<=m) c+=n;
      else if (a.at(i)>0) {
        if (a.at(i)*a.at(0)>m) continue;
        ll=0;rr=n;
        while (rr-ll>1) {
          mm=(ll+rr)/2;
          if (a.at(i)*a.at(mm)<=m) ll=mm;
          else rr=mm;
        }
        c+=ll+1;
      }
      if (a.at(i)*a.at(i)<=m) c--;
    }
    if (c<2*k) l=m;
    else r=m;
  }
  cout<<r<<endl;
}
```