#include <iostream>
#include <vector>
#include <algorithm>

typedef long long ll;

int main(){
   ll N, K;
   std::cin >> N >> K;
   std::vector<ll> A(N);
   for(int i=0; i<N; ++i){
      std::cin >> A[i];
   }
   std::sort(A.begin(), A.end());

   ll ulim = 2e18;
   ll llim = -2e18;
   while(ulim-llim>1){
      ll x = (ulim + llim)/2;
      ll count = 0;
      for(int i=0; i<N; ++i){
	 if(A[i]<0) count += N - (std::upper_bound(A.begin(), A.end(), (long double)x/A[i]) - A.begin()); 
	 else if(A[i]>0) count += std::lower_bound(A.begin(), A.end(), (long double)x/A[i]) - A.begin();
	 else if(x>0) count += N;
	 if(A[i]*A[i] < x) --count;
      }
      (count/2 < K ? llim : ulim) = x;
   }
   std::cout << llim << std::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 <iostream>
#include <vector>
#include <algorithm>

typedef long long ll;

int main(){
   ll N, K;
   std::cin >> N >> K;
   std::vector<ll> A(N);
   for(int i=0; i<N; ++i){
      std::cin >> A[i];
   }
   std::sort(A.begin(), A.end());

   ll ulim = 2e18;
   ll llim = -2e18;
   while(ulim-llim>1){
      ll x = (ulim + llim)/2;
      ll count = 0;
      for(int i=0; i<N; ++i){
	 if(A[i]<0) count += N - (std::upper_bound(A.begin(), A.end(), (long double)x/A[i]) - A.begin()); 
	 else if(A[i]>0) count += std::lower_bound(A.begin(), A.end(), (long double)x/A[i]) - A.begin();
	 else if(x>0) count += N;
	 if(A[i]*A[i] < x) --count;
      }
      (count/2 < K ? llim : ulim) = x;
   }
   std::cout << llim << std::endl;
   return 0;
}

```