#include <bits/stdc++.h> 

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

int main() {
	int n = ri();
	int64_t k;
	std::cin >> k;
	k--;
	k *= 2;
	int a[n];
	for (auto &i : a) i = ri();
	std::sort(a, a + n);
	int64_t l = -1000000000000000001, r = 1000000000000000001;
	while (r - l > 1) {
		int64_t m = l + (r - l) / 2;
		int64_t cnt = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] > 0) cnt += std::lower_bound(a, a + n, a[i], [&] (int i, int j) { return (int64_t) i * j <= m; }) - a;
			else if (a[i] < 0) cnt += a + n - std::lower_bound(a, a + n, a[i], [&] (int i, int j) { return (int64_t) i * j > m; });
			else cnt += m >= 0 ? n : 0;
			if ((int64_t) a[i] * a[i] <= m) cnt--;
		}
		std::cerr << m << ":" << cnt << std::endl;
		if (cnt <= k) l = m;
		else r = m;
	}
	printf("%lld\n", (long long) r);
	return 0;
}

### Prompt

Please provide a Cpp coded solution 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> 

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

int main() {
	int n = ri();
	int64_t k;
	std::cin >> k;
	k--;
	k *= 2;
	int a[n];
	for (auto &i : a) i = ri();
	std::sort(a, a + n);
	int64_t l = -1000000000000000001, r = 1000000000000000001;
	while (r - l > 1) {
		int64_t m = l + (r - l) / 2;
		int64_t cnt = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] > 0) cnt += std::lower_bound(a, a + n, a[i], [&] (int i, int j) { return (int64_t) i * j <= m; }) - a;
			else if (a[i] < 0) cnt += a + n - std::lower_bound(a, a + n, a[i], [&] (int i, int j) { return (int64_t) i * j > m; });
			else cnt += m >= 0 ? n : 0;
			if ((int64_t) a[i] * a[i] <= m) cnt--;
		}
		std::cerr << m << ":" << cnt << std::endl;
		if (cnt <= k) l = m;
		else r = m;
	}
	printf("%lld\n", (long long) r);
	return 0;
}

```