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

using namespace std;

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

	sort(a.begin(), a.end());
	
	Int low = -1e18 - 1, high = 1e18;
	while (high - low > 1) {
		Int mid = (high + low) / 2;

		Int cnt = 0;
		for (int i = 0; i + 1 < N; ++i) {
			if (a[i] >= 0) {
				if (a[i] * a[i + 1] > mid) { continue; }
				int l = i + 1, h = N;
				while (h - l > 1) {
					int m = (h + l) / 2;
					if (a[i] * a[m] > mid) { h = m; }
					else { l = m; }
				}
				cnt += h - i - 1;
			} else {
				if (a[i] * a.back() > mid) { continue; }
				int l = i, h = N - 1;
				while (h - l > 1) {
					int m = (h + l) / 2;
					if (a[i] * a[m] > mid) { l = m; }
					else { h = m; }
				}
				cnt += N - 1 - l;
			}
		}
		if (cnt >= K) { high = mid; }
		else { low = mid; }
	}
	cout << high << endl;

	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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

	sort(a.begin(), a.end());
	
	Int low = -1e18 - 1, high = 1e18;
	while (high - low > 1) {
		Int mid = (high + low) / 2;

		Int cnt = 0;
		for (int i = 0; i + 1 < N; ++i) {
			if (a[i] >= 0) {
				if (a[i] * a[i + 1] > mid) { continue; }
				int l = i + 1, h = N;
				while (h - l > 1) {
					int m = (h + l) / 2;
					if (a[i] * a[m] > mid) { h = m; }
					else { l = m; }
				}
				cnt += h - i - 1;
			} else {
				if (a[i] * a.back() > mid) { continue; }
				int l = i, h = N - 1;
				while (h - l > 1) {
					int m = (h + l) / 2;
					if (a[i] * a[m] > mid) { l = m; }
					else { h = m; }
				}
				cnt += N - 1 - l;
			}
		}
		if (cnt >= K) { high = mid; }
		else { low = mid; }
	}
	cout << high << endl;

	return 0;
}

```