#include <iostream>
#include <algorithm>
using namespace std;

int N; long long A[200200], R[200200], K;

int main()
{
	cin >> N >> K; K *= 2;
	for (int i = 0; i < N; i++){
		cin >> A[i];
		R[i] = -A[i];
	}
	sort(A, A + N);
	sort(R, R + N);

	long long l = -1e18, r = 1e18;
	while (l + 1 < r){
		long long m = l + (r - l) / 2, c = 0;
		for (int i = 0; i < N; i++){
			if (A[i] == 0){
				if (m >= 0) c += N - 1;
			}
			else if (A[i] > 0){
				long long u = (m >= 0 ? m / A[i] : (m - A[i] + 1) / A[i]);
				c += upper_bound(A, A + N, u) - A;
				if (A[i] * A[i] <= m) c--;
			}
			else{
				long long u = (m >= 0 ? m / (-A[i]) : (m + A[i] + 1) / (-A[i]));
				c += upper_bound(R, R + N, u) - R;
				if (A[i] * A[i] <= m) c--;
			}
		}

		if (c < K) l = m;
		else r = m;
	}

	cout << r << endl;

	return 0;
}

### Prompt

In cpp, your task is to solve 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 <algorithm>
using namespace std;

int N; long long A[200200], R[200200], K;

int main()
{
	cin >> N >> K; K *= 2;
	for (int i = 0; i < N; i++){
		cin >> A[i];
		R[i] = -A[i];
	}
	sort(A, A + N);
	sort(R, R + N);

	long long l = -1e18, r = 1e18;
	while (l + 1 < r){
		long long m = l + (r - l) / 2, c = 0;
		for (int i = 0; i < N; i++){
			if (A[i] == 0){
				if (m >= 0) c += N - 1;
			}
			else if (A[i] > 0){
				long long u = (m >= 0 ? m / A[i] : (m - A[i] + 1) / A[i]);
				c += upper_bound(A, A + N, u) - A;
				if (A[i] * A[i] <= m) c--;
			}
			else{
				long long u = (m >= 0 ? m / (-A[i]) : (m + A[i] + 1) / (-A[i]));
				c += upper_bound(R, R + N, u) - R;
				if (A[i] * A[i] <= m) c--;
			}
		}

		if (c < K) l = m;
		else r = m;
	}

	cout << r << endl;

	return 0;
}

```