#include<bits/stdc++.h>
using namespace std;
long long n,k,a[1000005],L=-1e18,R=1e18,mod=1e9+7,Mid;
long long check(long long x){
	long long res=0;
	for(int i=0;i<n;i++)
	{
		if(a[i]<0)
		{
			long long L=i,R=n;
			while(R-L>1)
			{
				long long mid=L+R>>1;
				if(a[i]*a[mid]<=x) R=mid;
				else L=mid;
			}
			res+=n-R;
		}
		else
		{
			long long L=i,R=n;
			while(R-L>1)
			{
				long long mid=L+R>>1;
				if(a[i]*a[mid]<=x) L=mid;
				else R=mid;
			}
			res+=L-i;
		}
	}
	return res;
}
int main(){
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a,a+n);
	while(R-L>1)
	{
		Mid=L+R>>1;
		if(check(Mid)<k) L=Mid;
		else R=Mid;
	}
	cout<<R;
	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;
long long n,k,a[1000005],L=-1e18,R=1e18,mod=1e9+7,Mid;
long long check(long long x){
	long long res=0;
	for(int i=0;i<n;i++)
	{
		if(a[i]<0)
		{
			long long L=i,R=n;
			while(R-L>1)
			{
				long long mid=L+R>>1;
				if(a[i]*a[mid]<=x) R=mid;
				else L=mid;
			}
			res+=n-R;
		}
		else
		{
			long long L=i,R=n;
			while(R-L>1)
			{
				long long mid=L+R>>1;
				if(a[i]*a[mid]<=x) L=mid;
				else R=mid;
			}
			res+=L-i;
		}
	}
	return res;
}
int main(){
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a,a+n);
	while(R-L>1)
	{
		Mid=L+R>>1;
		if(check(Mid)<k) L=Mid;
		else R=Mid;
	}
	cout<<R;
	return 0;
}
```