#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;

const ll INF = ll(1e18)+1;

ll N, K;
ll A[1000000];

ll calcRank(ll x) {
  ll totalCount = 0;
  for (int i = 0; i < N; i++) {
    if(A[i] >= 0){
      int l = -1;
      int r = N;
      while (l+1<r) {
        int mid = (l+r)/2;

        if (A[mid]*A[i] < x) l=mid;
        else r=mid;
      }
      totalCount += r;
    } else {
      int l = -1;
      int r = N;
      while (l+1<r) {
        int mid = (l+r)/2;
        if (A[mid]*A[i] < x) r=mid;
        else l=mid;
      }
      totalCount += N-r;
    }
  }
  for (int i = 0; i < N; i++) {
    if (A[i]*A[i] < x) totalCount--;
  }    
  totalCount /=2;
  return totalCount;
}



int main(){
  cin >> N >> K;
  
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  sort(A, A+N);
  
  ll l = -INF;
  ll r = INF;
  while (l+1 < r) {
    ll mid = (l+r)/2;   
    
    if (calcRank(mid)<K) l = mid;
    else r = mid;
  }
  
  cout << l << endl;
  return 0;
}
### 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;
using ll = long long;
using pii = pair<int, int>;

const ll INF = ll(1e18)+1;

ll N, K;
ll A[1000000];

ll calcRank(ll x) {
  ll totalCount = 0;
  for (int i = 0; i < N; i++) {
    if(A[i] >= 0){
      int l = -1;
      int r = N;
      while (l+1<r) {
        int mid = (l+r)/2;

        if (A[mid]*A[i] < x) l=mid;
        else r=mid;
      }
      totalCount += r;
    } else {
      int l = -1;
      int r = N;
      while (l+1<r) {
        int mid = (l+r)/2;
        if (A[mid]*A[i] < x) r=mid;
        else l=mid;
      }
      totalCount += N-r;
    }
  }
  for (int i = 0; i < N; i++) {
    if (A[i]*A[i] < x) totalCount--;
  }    
  totalCount /=2;
  return totalCount;
}



int main(){
  cin >> N >> K;
  
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  sort(A, A+N);
  
  ll l = -INF;
  ll r = INF;
  while (l+1 < r) {
    ll mid = (l+r)/2;   
    
    if (calcRank(mid)<K) l = mid;
    else r = mid;
  }
  
  cout << l << endl;
  return 0;
}
```