/*
* @author:  codancer
* @createTime:  2020-02-17, 09:39:06
*/
#include<bits/stdc++.h>
 
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
#define pb push_back
#define SZ(x) ((int)(x).size())
#define fi first
#define se second
typedef vector<int> VI;
typedef pair<int,int> pii;
ll n,k;
ll a[200010];
bool check(ll x){
	ll cnt=0;// the number that less than x
	for(int i=1;i<=n;i++){
		ll num=0;
		if(a[i]==0){
			if(x>0) num=n;
		}
		else if(a[i]>0){
			ll h=x/a[i];
			if(x%a[i]&&(x>0)){
				++h;
			}
			num=lower_bound(a+1,a+n+1,h)-a-1;
		}
		else{
			ll h=x/a[i];
			if(x%a[i]&&(x>0)){
				h--;
			}
			ll great=upper_bound(a+1,a+n+1,h)-a-1;
			num=n-great;
		}
		cnt+=num;
	}
	for(int i=1;i<=n;i++){
		if(a[i]*a[i]<x) --cnt;
	}
	cnt/=2;
	return cnt<k;
}
int main(){
	cin>>n>>k;
	for(int i=1;i<=n;i++) cin>>a[i];
	sort(a+1,a+n+1);
	ll l=-3e18,r=3e18;
	// cout<<check(-6)<<endl;
	for(int i=1;i<=100;i++){
		ll mid=(l+r)/2;
		if(check(mid)) l=mid;
		else r=mid;
	}
	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
/*
* @author:  codancer
* @createTime:  2020-02-17, 09:39:06
*/
#include<bits/stdc++.h>
 
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;
#define pb push_back
#define SZ(x) ((int)(x).size())
#define fi first
#define se second
typedef vector<int> VI;
typedef pair<int,int> pii;
ll n,k;
ll a[200010];
bool check(ll x){
	ll cnt=0;// the number that less than x
	for(int i=1;i<=n;i++){
		ll num=0;
		if(a[i]==0){
			if(x>0) num=n;
		}
		else if(a[i]>0){
			ll h=x/a[i];
			if(x%a[i]&&(x>0)){
				++h;
			}
			num=lower_bound(a+1,a+n+1,h)-a-1;
		}
		else{
			ll h=x/a[i];
			if(x%a[i]&&(x>0)){
				h--;
			}
			ll great=upper_bound(a+1,a+n+1,h)-a-1;
			num=n-great;
		}
		cnt+=num;
	}
	for(int i=1;i<=n;i++){
		if(a[i]*a[i]<x) --cnt;
	}
	cnt/=2;
	return cnt<k;
}
int main(){
	cin>>n>>k;
	for(int i=1;i<=n;i++) cin>>a[i];
	sort(a+1,a+n+1);
	ll l=-3e18,r=3e18;
	// cout<<check(-6)<<endl;
	for(int i=1;i<=100;i++){
		ll mid=(l+r)/2;
		if(check(mid)) l=mid;
		else r=mid;
	}
	cout<<l<<endl;
	return 0;
}
```