#include<bits/stdc++.h>
using namespace std;
int n;
long long k,cl,cr,cm,l,r,m,tot,a[2000100],INF=1e9+1;
long long Get(){
	for(int i=0;i<n;i++){
		if(a[i]<0){
			l=-1,r=n;
			while(l+1<r){
				m=(l+r)/2LL;
				if(a[i]*a[m]<cm) r=m;
				else l=m;
			}
			tot+=(long long)(n)-r;
		}
		else{
			l=-1,r=n;
			while(l+1<r){
				m=(l+r)/2LL;
				if(a[i]*a[m]<cm) l=m;
				else r=m;
			}
			tot+=r;
		}
		if((a[i]*a[i])<cm) tot-=1LL;
	}
	tot/=2LL;
}
int main(){
	#ifdef FILIN
		#ifndef DavidDesktop
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>k;
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
	INF*=INF,cl=-INF,cr=INF;
	while(cl+1<cr){
		cm=(cl+cr)/2LL,tot=0LL;
		if(Get()<k) cl=cm;
		else cr=cm;
		if(tot<k) cl=cm;
		else cr=cm;
	}
	cout<<cl<<endl;
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int n;
long long k,cl,cr,cm,l,r,m,tot,a[2000100],INF=1e9+1;
long long Get(){
	for(int i=0;i<n;i++){
		if(a[i]<0){
			l=-1,r=n;
			while(l+1<r){
				m=(l+r)/2LL;
				if(a[i]*a[m]<cm) r=m;
				else l=m;
			}
			tot+=(long long)(n)-r;
		}
		else{
			l=-1,r=n;
			while(l+1<r){
				m=(l+r)/2LL;
				if(a[i]*a[m]<cm) l=m;
				else r=m;
			}
			tot+=r;
		}
		if((a[i]*a[i])<cm) tot-=1LL;
	}
	tot/=2LL;
}
int main(){
	#ifdef FILIN
		#ifndef DavidDesktop
			freopen(FILIN,"r",stdin);
			freopen(FILOUT,"w",stdout);
		#endif
	#endif
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>k;
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
	INF*=INF,cl=-INF,cr=INF;
	while(cl+1<cr){
		cm=(cl+cr)/2LL,tot=0LL;
		if(Get()<k) cl=cm;
		else cr=cm;
		if(tot<k) cl=cm;
		else cr=cm;
	}
	cout<<cl<<endl;
	return 0;
}

```