#include<bits/stdc++.h>
#define ll long long 
using namespace std;
int main(){
	ios_base::sync_with_stdio(0) ; 
	cin.tie(0) ; 
	ll n , k  ;
	cin>> n >> k  ;
	vector<ll> a(n) ;
//	ll neg = 0 , pos = 0 , zero = 0 ;  
	for(ll &x : a) {
		cin>> x ; 
		
	}
	sort(a.begin() , a.end()) ; 
	ll p1 = -1e18 ; 
	ll p2 = 1e18 ; 
   while(p2 !=  p1 +  1){
	   ll mid  = (p1 + p2)/2 ; 
	   ll cnt = 0 ; 
	   for(int i = 0 ; i < n ; ++i){
		   if(a[i] < 0 ){
			   ll l = i  , en = n ; 
			   while(l + 1 != en){
				   ll m = (l + en)/2 ; 
				   if(a[i]*a[m] <= mid) en = m ; 
				   else l = m ; 
			   }
			   cnt+=(n-en) ; 
			   
		   }
		   else{
			   ll l = i  , en = n ; 
			   while( l + 1 != en ){
				  ll  m = (l + en)/2 ; 
				  if(a[i]*a[m] <= mid) l = m ; 
				  else en = m ; 
			  }
			  cnt+=(l -i );    
			   
		   }
	  
}
  if(cnt <  k){
	      p1 = mid ; 
	  }
	      else 
	      p2 = mid ; 
	   
   } 
  cout<<p2<<endl;
 
 
 
}


### 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<bits/stdc++.h>
#define ll long long 
using namespace std;
int main(){
	ios_base::sync_with_stdio(0) ; 
	cin.tie(0) ; 
	ll n , k  ;
	cin>> n >> k  ;
	vector<ll> a(n) ;
//	ll neg = 0 , pos = 0 , zero = 0 ;  
	for(ll &x : a) {
		cin>> x ; 
		
	}
	sort(a.begin() , a.end()) ; 
	ll p1 = -1e18 ; 
	ll p2 = 1e18 ; 
   while(p2 !=  p1 +  1){
	   ll mid  = (p1 + p2)/2 ; 
	   ll cnt = 0 ; 
	   for(int i = 0 ; i < n ; ++i){
		   if(a[i] < 0 ){
			   ll l = i  , en = n ; 
			   while(l + 1 != en){
				   ll m = (l + en)/2 ; 
				   if(a[i]*a[m] <= mid) en = m ; 
				   else l = m ; 
			   }
			   cnt+=(n-en) ; 
			   
		   }
		   else{
			   ll l = i  , en = n ; 
			   while( l + 1 != en ){
				  ll  m = (l + en)/2 ; 
				  if(a[i]*a[m] <= mid) l = m ; 
				  else en = m ; 
			  }
			  cnt+=(l -i );    
			   
		   }
	  
}
  if(cnt <  k){
	      p1 = mid ; 
	  }
	      else 
	      p2 = mid ; 
	   
   } 
  cout<<p2<<endl;
 
 
 
}


```