#include <bits/stdc++.h>
#define vl vector<ll>
using namespace std;
typedef long long ll;
ll div_ceil(ll a,ll b)
{
	if(a>=0)
		return (a+b-1)/b;
	else
		return a/b;
}
ll div_floor(ll a,ll b)
{
	if(a<0)
		return (a-b+1)/b;
	else
		return a/b;
}
int main()
{
	ll n,k;
	cin>>n>>k;
	vl A(n);
	for(int i=0;i<n;i++)cin>>A[i];
	ll inf = 1e18;
	ll ri = inf,le = -inf-1;
	sort(A.begin(),A.end());
//	(le , ri]
	while(le+1<ri)
	{
		ll mid = (le+ri) >> 1;
		ll cnt = 0;
		for(int i=0;i<n;i++)
		{
			if(A[i]<0)
			{
				ll num = div_ceil(-mid,-A[i]);
				cnt+=A.begin()+i - lower_bound(A.begin(),A.begin()+i,num);
			}
			else if(A[i] == 0)
			{
				if(mid>=0)
					cnt+=i;
			}
			else
			{
				ll num = div_floor(mid,A[i]);
				cnt += upper_bound(A.begin(),A.begin()+i,num)-A.begin();
			}
		}
		if(cnt>=k)
			ri = mid;
		else
			le = mid;
	}
	cout<<ri<<endl;
    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>
#define vl vector<ll>
using namespace std;
typedef long long ll;
ll div_ceil(ll a,ll b)
{
	if(a>=0)
		return (a+b-1)/b;
	else
		return a/b;
}
ll div_floor(ll a,ll b)
{
	if(a<0)
		return (a-b+1)/b;
	else
		return a/b;
}
int main()
{
	ll n,k;
	cin>>n>>k;
	vl A(n);
	for(int i=0;i<n;i++)cin>>A[i];
	ll inf = 1e18;
	ll ri = inf,le = -inf-1;
	sort(A.begin(),A.end());
//	(le , ri]
	while(le+1<ri)
	{
		ll mid = (le+ri) >> 1;
		ll cnt = 0;
		for(int i=0;i<n;i++)
		{
			if(A[i]<0)
			{
				ll num = div_ceil(-mid,-A[i]);
				cnt+=A.begin()+i - lower_bound(A.begin(),A.begin()+i,num);
			}
			else if(A[i] == 0)
			{
				if(mid>=0)
					cnt+=i;
			}
			else
			{
				ll num = div_floor(mid,A[i]);
				cnt += upper_bound(A.begin(),A.begin()+i,num)-A.begin();
			}
		}
		if(cnt>=k)
			ri = mid;
		else
			le = mid;
	}
	cout<<ri<<endl;
    return 0;
} 
```