#include<bits/stdc++.h>

using namespace std;

const long long INF = (1ll << 60);

bool many(long long m, long long K, vector<long long>&A){
  int N = A.size();
  for(int i = 0; i < N; i++){
    if(A[i] >= 0){
      int ok = i, ng = N;
      while(ng - ok > 1){
        int mid = (ok + ng) / 2;
        (A[i] * A[mid] <= m ? ok : ng) = mid;
      }
      K -= ok - i;
    }else{
      int ok = N, ng = i;
      while(ok - ng > 1){
        int mid = (ok + ng) / 2;
        (A[i] * A[mid] <= m ? ok : ng) = mid;
      }
      K -= N - ok;
    }
  }
  
  return K <= 0;
}

int main(){
  int N;
  long long K;
  cin >> N >> K;
  
  vector<long long> A(N);
  for(int i = 0; i < N; i++) cin >> A[i];
  sort(A.begin(), A.end());
  
  long long ok = INF, ng = -INF;
  while(ok - ng > 1){
    long long mid = (ok + ng) / 2;
    (many(mid, K, A) ? ok : ng) = mid;
  }
  
  cout << ok << endl;
  
  return 0;
}
### Prompt

Please create a solution in Cpp 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;

const long long INF = (1ll << 60);

bool many(long long m, long long K, vector<long long>&A){
  int N = A.size();
  for(int i = 0; i < N; i++){
    if(A[i] >= 0){
      int ok = i, ng = N;
      while(ng - ok > 1){
        int mid = (ok + ng) / 2;
        (A[i] * A[mid] <= m ? ok : ng) = mid;
      }
      K -= ok - i;
    }else{
      int ok = N, ng = i;
      while(ok - ng > 1){
        int mid = (ok + ng) / 2;
        (A[i] * A[mid] <= m ? ok : ng) = mid;
      }
      K -= N - ok;
    }
  }
  
  return K <= 0;
}

int main(){
  int N;
  long long K;
  cin >> N >> K;
  
  vector<long long> A(N);
  for(int i = 0; i < N; i++) cin >> A[i];
  sort(A.begin(), A.end());
  
  long long ok = INF, ng = -INF;
  while(ok - ng > 1){
    long long mid = (ok + ng) / 2;
    (many(mid, K, A) ? ok : ng) = mid;
  }
  
  cout << ok << endl;
  
  return 0;
}
```