#include <bits/stdc++.h>

using namespace std;

#ifdef DGC
#include "debug.h"
#else
#define debug(...) 9715
#endif
typedef long long ll;
typedef long double ld;
typedef complex<ll> point;
#define F first
#define S second



int main()
{
	#ifdef DGC
		//freopen("a.in", "r", stdin);
		//freopen("b.out", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	ll n, k;
	cin >> n >> k;
	vector<ll> a(n);
	for (auto &i : a) cin >> i;
	sort(a.begin(), a.end());

	auto f = [&](ld x)
	{
		ll r = 0;
		for (auto i : a)
		{
			if (i == 0) r += n * (x >= 0);
			else
			{
				ll x1 = floorl(x / i);
				ll x2 = ceill(x / i);
				if (i > 0) r += upper_bound(a.begin(), a.end(), x1) - a.begin();
				else r += a.end() - lower_bound(a.begin(), a.end(), x2);
			}
			r -= i*i <= x;
		}
		return r / 2;
	};

	ll l = -1e18, r = 1e18;
	while (l < r)
	{
		ll md = (l + r) >> 1;
		if (f(md) < k)
			l = md + 1;
		else
			r = md;
	}
	cout << l << "\n";

	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 <bits/stdc++.h>

using namespace std;

#ifdef DGC
#include "debug.h"
#else
#define debug(...) 9715
#endif
typedef long long ll;
typedef long double ld;
typedef complex<ll> point;
#define F first
#define S second



int main()
{
	#ifdef DGC
		//freopen("a.in", "r", stdin);
		//freopen("b.out", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	ll n, k;
	cin >> n >> k;
	vector<ll> a(n);
	for (auto &i : a) cin >> i;
	sort(a.begin(), a.end());

	auto f = [&](ld x)
	{
		ll r = 0;
		for (auto i : a)
		{
			if (i == 0) r += n * (x >= 0);
			else
			{
				ll x1 = floorl(x / i);
				ll x2 = ceill(x / i);
				if (i > 0) r += upper_bound(a.begin(), a.end(), x1) - a.begin();
				else r += a.end() - lower_bound(a.begin(), a.end(), x2);
			}
			r -= i*i <= x;
		}
		return r / 2;
	};

	ll l = -1e18, r = 1e18;
	while (l < r)
	{
		ll md = (l + r) >> 1;
		if (f(md) < k)
			l = md + 1;
		else
			r = md;
	}
	cout << l << "\n";

	return 0;
}

```