#include <bits/stdc++.h>
using namespace std;
int main(){
	int N; long long K;
	cin >> N >> K; 
	long long A[200010], S[200010] = {}; // Sにはヘイホーを入れる
	for(int i = 0; i < N; i++){
		cin >> A[i]; 
		S[i] = A[i] * A[i]; 
	}
	sort(A, A+N); 
	sort(S, S+N); 
	long long ng = -1e18-7; 
	long long ok = 1e18+7; 
	K *= 2; K--; // 重複を考慮すると2K-1番目。ヘイホーはにぶたんの中で処理
	while(ok - ng > 1){
		long long mid = (ok + ng) / 2; 
		long long cnt = 0; 
		long long M = K; 
		M += upper_bound(S, S+N, mid) - S; // mid以下のものがM個以上か
		for(int i = 0; i < N; i++){
			if(A[i] == 0 && mid >= 0) cnt += N; 
			else if(A[i] < 0){
				cnt += N - (lower_bound(A, A+N, (long double) mid / A[i]) - A); // あやしい
			}
			else{
				cnt += upper_bound(A, A+N, (long double) mid / A[i]) - A; // あやしい
			}
		}
		if(cnt < M) ng = mid; 
		else ok = mid;  
	}
	cout << ok << endl; 
}
### Prompt

Create a solution in cpp for 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(){
	int N; long long K;
	cin >> N >> K; 
	long long A[200010], S[200010] = {}; // Sにはヘイホーを入れる
	for(int i = 0; i < N; i++){
		cin >> A[i]; 
		S[i] = A[i] * A[i]; 
	}
	sort(A, A+N); 
	sort(S, S+N); 
	long long ng = -1e18-7; 
	long long ok = 1e18+7; 
	K *= 2; K--; // 重複を考慮すると2K-1番目。ヘイホーはにぶたんの中で処理
	while(ok - ng > 1){
		long long mid = (ok + ng) / 2; 
		long long cnt = 0; 
		long long M = K; 
		M += upper_bound(S, S+N, mid) - S; // mid以下のものがM個以上か
		for(int i = 0; i < N; i++){
			if(A[i] == 0 && mid >= 0) cnt += N; 
			else if(A[i] < 0){
				cnt += N - (lower_bound(A, A+N, (long double) mid / A[i]) - A); // あやしい
			}
			else{
				cnt += upper_bound(A, A+N, (long double) mid / A[i]) - A; // あやしい
			}
		}
		if(cnt < M) ng = mid; 
		else ok = mid;  
	}
	cout << ok << endl; 
}
```