#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

int main()
{
	long long n, k;
	scanf("%lld%lld", &n, &k);
	long long a[200000];
	for (long long i = 0; i < n; ++i)
	{
		scanf("%lld", a + i);
	}
	std::sort(a, a + n);
	long long l = -1000000000000000001;
	long long r = 1000000000000000001;
	while (r - l > 1)
	{
		long long mid = l + (r - l) / 2;
		long long cnt = 0;
		for (long long i = 0; i < n - 1; ++i)
		{
			long long search;
			if (a[i] < 0)
			{
				search = mid <= 0 ? (-mid + (-a[i] - 1)) / (-a[i]) : mid / a[i];
				cnt += (a + n) - std::lower_bound(a + i + 1, a + n, search);
			}
			else if (a[i] == 0)
			{
				if (mid >= 0) cnt += n - (i + 1);
			}
			else
			{
				search = mid <= 0 ? (mid - (a[i] - 1)) / a[i] : mid / a[i];
				cnt += std::upper_bound(a + i + 1, a + n, search) - (a + i + 1);
			}
		}
		if (cnt >= k)
		{
			r = mid;
		}
		else
		{
			l = mid;
		}
	}
	printf("%lld\n", r);
}

### Prompt

Please create a solution in cpp 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 <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

int main()
{
	long long n, k;
	scanf("%lld%lld", &n, &k);
	long long a[200000];
	for (long long i = 0; i < n; ++i)
	{
		scanf("%lld", a + i);
	}
	std::sort(a, a + n);
	long long l = -1000000000000000001;
	long long r = 1000000000000000001;
	while (r - l > 1)
	{
		long long mid = l + (r - l) / 2;
		long long cnt = 0;
		for (long long i = 0; i < n - 1; ++i)
		{
			long long search;
			if (a[i] < 0)
			{
				search = mid <= 0 ? (-mid + (-a[i] - 1)) / (-a[i]) : mid / a[i];
				cnt += (a + n) - std::lower_bound(a + i + 1, a + n, search);
			}
			else if (a[i] == 0)
			{
				if (mid >= 0) cnt += n - (i + 1);
			}
			else
			{
				search = mid <= 0 ? (mid - (a[i] - 1)) / a[i] : mid / a[i];
				cnt += std::upper_bound(a + i + 1, a + n, search) - (a + i + 1);
			}
		}
		if (cnt >= k)
		{
			r = mid;
		}
		else
		{
			l = mid;
		}
	}
	printf("%lld\n", r);
}

```