
#include <iostream>
#include<map>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define mset(s,_) memset(s,_,sizeof s)
using namespace std;
ll a[200010];
int main(){
	ll n,k;
	cin>>n>>k;
	rep(i,0,n-1) cin>>a[i];
	ll l=-1e18,r=1e18;
	sort(a,a+n);
	while(l<r){
		ll mid=(l+r)>>1,cnt=0;
		rep(i,0,n-1){
			if(a[i]>0){
				cnt+=upper_bound(a,a+n,(long double)mid/a[i])-a;
		  }
		  else if(a[i]<0){
		  	cnt+=n-(lower_bound(a,a+n,(long double)mid/a[i])-a);
		  }
		  else if(mid>=0){
		  	cnt+=n;
		  }
		  if(a[i]*a[i]<=mid) cnt--;
		}
		if(cnt>=2*k) r=mid;
    else l=mid+1;
   //cout<<mid<<'\n';
	}
	cout<<l<<'\n';
}
			
	
### Prompt

Please formulate 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 <iostream>
#include<map>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define mset(s,_) memset(s,_,sizeof s)
using namespace std;
ll a[200010];
int main(){
	ll n,k;
	cin>>n>>k;
	rep(i,0,n-1) cin>>a[i];
	ll l=-1e18,r=1e18;
	sort(a,a+n);
	while(l<r){
		ll mid=(l+r)>>1,cnt=0;
		rep(i,0,n-1){
			if(a[i]>0){
				cnt+=upper_bound(a,a+n,(long double)mid/a[i])-a;
		  }
		  else if(a[i]<0){
		  	cnt+=n-(lower_bound(a,a+n,(long double)mid/a[i])-a);
		  }
		  else if(mid>=0){
		  	cnt+=n;
		  }
		  if(a[i]*a[i]<=mid) cnt--;
		}
		if(cnt>=2*k) r=mid;
    else l=mid+1;
   //cout<<mid<<'\n';
	}
	cout<<l<<'\n';
}
			
	
```