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

using namespace std;
typedef long long ll;

const ll INF = ll(1e18) + 1;

int main(int argc, char *argv[])
{
	int N;
	ll K;

	cin >> N >> K;

	vector<ll> a(N);
	for (int i = 0; i < N; i++) {
		cin >> a[i];
	}
	sort(a.begin(), a.end());
	ll l = -INF, r = INF;
	while (l + 1 < r) {
		ll x = (l + r) / 2;
		bool ok = [&]{
			ll tot = 0;
			for (int i = 0; i < N; i++) {
				if (a[i] < 0) {
					int l = -1, r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if  (a[c] * a[i] < x) {
							r = c;
						} else {
							l = c;
						}
					}
					tot += N - r;
				} else {
					int l = -1, r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if (a[c] * a[i] < x) {
							l = c;
						} else {
							r = c;
						}
					}
					tot += r;
				}
				if (a[i] * a[i] < x) {
					tot--;
				}
			}
			tot /= 2;
			return tot < K;
		}();
		if (ok) {
			l = x;
		} else {
			r = x;
		}
	}
	cout << l << endl;
	return 0;
}
### Prompt

Your task is to create a cpp solution 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 <algorithm>
#include <vector>

using namespace std;
typedef long long ll;

const ll INF = ll(1e18) + 1;

int main(int argc, char *argv[])
{
	int N;
	ll K;

	cin >> N >> K;

	vector<ll> a(N);
	for (int i = 0; i < N; i++) {
		cin >> a[i];
	}
	sort(a.begin(), a.end());
	ll l = -INF, r = INF;
	while (l + 1 < r) {
		ll x = (l + r) / 2;
		bool ok = [&]{
			ll tot = 0;
			for (int i = 0; i < N; i++) {
				if (a[i] < 0) {
					int l = -1, r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if  (a[c] * a[i] < x) {
							r = c;
						} else {
							l = c;
						}
					}
					tot += N - r;
				} else {
					int l = -1, r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if (a[c] * a[i] < x) {
							l = c;
						} else {
							r = c;
						}
					}
					tot += r;
				}
				if (a[i] * a[i] < x) {
					tot--;
				}
			}
			tot /= 2;
			return tot < K;
		}();
		if (ok) {
			l = x;
		} else {
			r = x;
		}
	}
	cout << l << endl;
	return 0;
}
```