#include<bits/stdc++.h>
using namespace std;
long long n,a[202020],ans,l,r,k;
bool check(long long x){
	long long sum=0;
	for(int i=0;i<n;i++){
		if(a[i]==0){
			if(x>=0){
				sum+=n-1;
			}
		}
		else if(a[i]<0){
			long long mx=x/a[i];
			if(mx*a[i]>x){
				mx++;
			}
			int j=lower_bound(a,a+n,mx)-a;
			sum+=n-j;
			if(j<=i){
				sum--;
			}
		}
		else{
			long long mx=x/a[i];
			if(mx*a[i]>x){
				mx--;
			}
			int j=upper_bound(a,a+n,mx)-a-1;
			if(j<0){
				continue;
			}
			sum+=j+1;
			if(j>=i){
				sum--;
			}
		}
	}
	if(sum/2>=k){
		return 1;
	}
	else return 0;
}
int main(){
	cin>>n>>k;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	sort(a,a+n);
	l=-1e18-100;
	r=1e18+100;
	while(l<r){
		long long mid=l+(r-l)/2;
		if(check(mid)){
			r=mid;
		}
		else{
			l=mid+1;
		}
	}
	cout<<l<<endl;
	return 0;
}
### Prompt

Generate 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,a[202020],ans,l,r,k;
bool check(long long x){
	long long sum=0;
	for(int i=0;i<n;i++){
		if(a[i]==0){
			if(x>=0){
				sum+=n-1;
			}
		}
		else if(a[i]<0){
			long long mx=x/a[i];
			if(mx*a[i]>x){
				mx++;
			}
			int j=lower_bound(a,a+n,mx)-a;
			sum+=n-j;
			if(j<=i){
				sum--;
			}
		}
		else{
			long long mx=x/a[i];
			if(mx*a[i]>x){
				mx--;
			}
			int j=upper_bound(a,a+n,mx)-a-1;
			if(j<0){
				continue;
			}
			sum+=j+1;
			if(j>=i){
				sum--;
			}
		}
	}
	if(sum/2>=k){
		return 1;
	}
	else return 0;
}
int main(){
	cin>>n>>k;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	sort(a,a+n);
	l=-1e18-100;
	r=1e18+100;
	while(l<r){
		long long mid=l+(r-l)/2;
		if(check(mid)){
			r=mid;
		}
		else{
			l=mid+1;
		}
	}
	cout<<l<<endl;
	return 0;
}
```