#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 1LL<<61;

int main(){
  ll n, k; cin >> n >> k;
  vector<ll> A(n);
  for(int i = 0; i < n; i++) cin >> A[i];
  sort(A.begin(), A.end());
  ll left = -INF, right = INF;
  while(right - left > 1){
    ll x = (right+left)/2;
    ll s = 0, t = 0;
    for(int i = 0; i < n; i++){
      if(A[i] > 0){
        int l1 = -1, r1 = n;
        while(r1 - l1 > 1){
          int m = (r1 + l1) / 2;
          if(A[i] * A[m] <= x) l1 = m; 
          else r1 = m;
        }
        s += r1;
      }
      else if(A[i] < 0){
        int l1 = -1, r1 = n;
        while(r1 - l1 > 1){
          int m = (r1 + l1) / 2;
          if(A[i] * A[m] <= x) r1 = m;
          else l1 = m;
        }
        s += n - r1;
      }
      else if(x >= 0) s += n;
      if(A[i] * A[i] <= x) t++;
    }
    ll num = (s - t) / 2;
    if(num >= k) right = x;
    else left = x;
  }
  cout <<  right << endl;
}
### 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;
using ll = long long;
const ll INF = 1LL<<61;

int main(){
  ll n, k; cin >> n >> k;
  vector<ll> A(n);
  for(int i = 0; i < n; i++) cin >> A[i];
  sort(A.begin(), A.end());
  ll left = -INF, right = INF;
  while(right - left > 1){
    ll x = (right+left)/2;
    ll s = 0, t = 0;
    for(int i = 0; i < n; i++){
      if(A[i] > 0){
        int l1 = -1, r1 = n;
        while(r1 - l1 > 1){
          int m = (r1 + l1) / 2;
          if(A[i] * A[m] <= x) l1 = m; 
          else r1 = m;
        }
        s += r1;
      }
      else if(A[i] < 0){
        int l1 = -1, r1 = n;
        while(r1 - l1 > 1){
          int m = (r1 + l1) / 2;
          if(A[i] * A[m] <= x) r1 = m;
          else l1 = m;
        }
        s += n - r1;
      }
      else if(x >= 0) s += n;
      if(A[i] * A[i] <= x) t++;
    }
    ll num = (s - t) / 2;
    if(num >= k) right = x;
    else left = x;
  }
  cout <<  right << endl;
}
```