#include <bits/stdc++.h>
using namespace std;
#define mp(a, b) make_pair(a, b)
#define lowbit(x) (x & (-x))
#define ls (rt << 1)
#define rs (rt << 1 | 1)
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 2e5 + 10;
ll n, k;
ll a[maxn];
ll solve(ll x)
{
	ll ret = 0;
	for(int i = 1; i <= n; ++i)
	{
		if(a[i] * a[i] <= x) --ret;
		if(a[i] > 0)
		{
			ll tmp = x / a[i];
			while(tmp * a[i] > x) --tmp;
			ret += upper_bound(a + 1, a + n + 1, tmp) - 1 - a;
		}
		else if(a[i] == 0)
		{
			if(x >= 0) ret += n;
		}
		else
		{
			ll tmp = x / a[i];
			while(tmp * a[i] > x) ++tmp;
			ret += a + n - lower_bound(a + 1, a + n + 1, tmp) + 1;
		}
	}
	return ret / 2;
}
int main()
{
	cin >> n >> k;
	for(int i = 1; i <= n; ++i)
		cin >> a[i];
	sort(a + 1, a + n + 1);	
	ll l = -1e18, r = 1e18;
	while(l <= r)
	{
		ll mid = (l + r) / 2;
		if(solve(mid) >= k)
			r = mid - 1;
		else
			l = mid + 1;
	}
	cout << l << endl;
	return 0;
}
/*
3 3
0 -1 1
*/
### 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>
using namespace std;
#define mp(a, b) make_pair(a, b)
#define lowbit(x) (x & (-x))
#define ls (rt << 1)
#define rs (rt << 1 | 1)
typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 2e5 + 10;
ll n, k;
ll a[maxn];
ll solve(ll x)
{
	ll ret = 0;
	for(int i = 1; i <= n; ++i)
	{
		if(a[i] * a[i] <= x) --ret;
		if(a[i] > 0)
		{
			ll tmp = x / a[i];
			while(tmp * a[i] > x) --tmp;
			ret += upper_bound(a + 1, a + n + 1, tmp) - 1 - a;
		}
		else if(a[i] == 0)
		{
			if(x >= 0) ret += n;
		}
		else
		{
			ll tmp = x / a[i];
			while(tmp * a[i] > x) ++tmp;
			ret += a + n - lower_bound(a + 1, a + n + 1, tmp) + 1;
		}
	}
	return ret / 2;
}
int main()
{
	cin >> n >> k;
	for(int i = 1; i <= n; ++i)
		cin >> a[i];
	sort(a + 1, a + n + 1);	
	ll l = -1e18, r = 1e18;
	while(l <= r)
	{
		ll mid = (l + r) / 2;
		if(solve(mid) >= k)
			r = mid - 1;
		else
			l = mid + 1;
	}
	cout << l << endl;
	return 0;
}
/*
3 3
0 -1 1
*/
```