#include<bits/stdc++.h>
using namespace std;
const int N=2e5;
int main()
{
	int n;
	long long k;
	ios::sync_with_stdio(false);
	cin>>n>>k;
	vector<long double>v;
	for(int i=1;i<=n;i++){
		int t;
		cin>>t;
		v.push_back(t);
	}
	sort(v.begin(),v.end());
	long long l=-1e18,r=1e18+1;
	auto check=[&](long long t){
		long long c=0;
		long double x=t;
		for(int i=0;i<n;i++){
			if(v[i]==0){
				if(x>=0)c+=i;
				continue;
			}
			if(v[i]>0){
				//t*x<=c;
				c+=upper_bound(v.begin(),v.begin()+i,x/v[i])-v.begin();
			}
			else {
				//t>=c/x
				c+=i-(lower_bound(v.begin(),v.begin()+i,x/v[i])-v.begin());
			}
		}
		return c>=k;
	};
	long long ans;
	while(l<r){
		long long mid=l+(r-l)/2;
		if(check(mid)){
			r=mid;
			ans=mid;
		}
		else {
			l=mid+1;
		}
	}
	cout<<ans<<'\n';
} 
### Prompt

Develop a solution in cpp 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;
const int N=2e5;
int main()
{
	int n;
	long long k;
	ios::sync_with_stdio(false);
	cin>>n>>k;
	vector<long double>v;
	for(int i=1;i<=n;i++){
		int t;
		cin>>t;
		v.push_back(t);
	}
	sort(v.begin(),v.end());
	long long l=-1e18,r=1e18+1;
	auto check=[&](long long t){
		long long c=0;
		long double x=t;
		for(int i=0;i<n;i++){
			if(v[i]==0){
				if(x>=0)c+=i;
				continue;
			}
			if(v[i]>0){
				//t*x<=c;
				c+=upper_bound(v.begin(),v.begin()+i,x/v[i])-v.begin();
			}
			else {
				//t>=c/x
				c+=i-(lower_bound(v.begin(),v.begin()+i,x/v[i])-v.begin());
			}
		}
		return c>=k;
	};
	long long ans;
	while(l<r){
		long long mid=l+(r-l)/2;
		if(check(mid)){
			r=mid;
			ans=mid;
		}
		else {
			l=mid+1;
		}
	}
	cout<<ans<<'\n';
} 
```