#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

int f(Int mid, Int x, vector<Int>& v) {
	if (x * v.back() <= mid) { return v.size(); }
	int ok = v.size(), ng = -1;
	while (ok - ng > 1) {
		int m = (ok + ng) / 2;
		(x * v[m] > mid ? ok : ng) = m;
	}
	return ok;
}

bool check(Int mid, Int K, vector<Int>& a, vector<Int>& ra) {
	Int cnt = 0;
	for (int i = 0; i < a.size(); ++i) {
		if (a[i] < 0) {
			cnt += f(mid, -a[i], ra);
		} else {
			cnt += f(mid, a[i], a);
		}
		cnt -= (a[i] * a[i] <= mid);
	}
	return cnt / 2 >= K;
}

int main() {
	Int N, K;
	cin >> N >> K;
	vector<Int> a(N), ra(N);
	for (int i = 0; i < N; ++i) {
		cin >> a[i];
		ra[i] = -a[i];
	}

	sort(a.begin(), a.end());
	sort(ra.begin(), ra.end());

	Int ok = 1e18, ng = -1e18 - 1;
	while (ok - ng > 1) {
		Int mid = (ok + ng) / 2;

		(check(mid, K, a, ra) ? ok : ng) = mid;
	}
	cout << ok << endl;

	return 0;
}

### 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>
#define Int int64_t

using namespace std;

int f(Int mid, Int x, vector<Int>& v) {
	if (x * v.back() <= mid) { return v.size(); }
	int ok = v.size(), ng = -1;
	while (ok - ng > 1) {
		int m = (ok + ng) / 2;
		(x * v[m] > mid ? ok : ng) = m;
	}
	return ok;
}

bool check(Int mid, Int K, vector<Int>& a, vector<Int>& ra) {
	Int cnt = 0;
	for (int i = 0; i < a.size(); ++i) {
		if (a[i] < 0) {
			cnt += f(mid, -a[i], ra);
		} else {
			cnt += f(mid, a[i], a);
		}
		cnt -= (a[i] * a[i] <= mid);
	}
	return cnt / 2 >= K;
}

int main() {
	Int N, K;
	cin >> N >> K;
	vector<Int> a(N), ra(N);
	for (int i = 0; i < N; ++i) {
		cin >> a[i];
		ra[i] = -a[i];
	}

	sort(a.begin(), a.end());
	sort(ra.begin(), ra.end());

	Int ok = 1e18, ng = -1e18 - 1;
	while (ok - ng > 1) {
		Int mid = (ok + ng) / 2;

		(check(mid, K, a, ra) ? ok : ng) = mid;
	}
	cout << ok << endl;

	return 0;
}

```