#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define pii pair<int,int>
#define two(x) (x<<1)
#define towd(x) (x>>1)
#define pb push_back
#define mp make_pair
#define sz(x) x.size()
#define x first
#define y second
 
ll a[200001];
 
int main(){
	std::ios::sync_with_stdio(false);
    std::cin.tie(0);
 
	ll n,k;
	cin>>n>>k;
	for (int i=0;i<n;i++) cin>>a[i];
	
	ll L=-2e18,R=2e18;
	sort(a,a+n);
	while (L+1!=R){
		ll m=towd(L+R);
		ll cnt=0;
		for (int i=0; i<n; i++){
			if (a[i]<0){
				ll l=i,r=n;
				while (l+1!=r){
					int mid=towd(l+r);
					if (a[mid]*a[i]<=m) r=mid;
					else l=mid;
				}
				cnt+=n-r;
			}
			else{
				ll l=i,r=n;
				while (l+1!=r){
					int mid=towd(l+r);
					if (a[mid]*a[i]<=m) l=mid;
					else r=mid;
				}
				cnt+=l-i;
			}
		}
		if (cnt<k) L=m;
		else R=m;
	}
	cout<<R<<endl;
	return 0;
}
### 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>
using namespace std;
 
#define ll long long
#define pii pair<int,int>
#define two(x) (x<<1)
#define towd(x) (x>>1)
#define pb push_back
#define mp make_pair
#define sz(x) x.size()
#define x first
#define y second
 
ll a[200001];
 
int main(){
	std::ios::sync_with_stdio(false);
    std::cin.tie(0);
 
	ll n,k;
	cin>>n>>k;
	for (int i=0;i<n;i++) cin>>a[i];
	
	ll L=-2e18,R=2e18;
	sort(a,a+n);
	while (L+1!=R){
		ll m=towd(L+R);
		ll cnt=0;
		for (int i=0; i<n; i++){
			if (a[i]<0){
				ll l=i,r=n;
				while (l+1!=r){
					int mid=towd(l+r);
					if (a[mid]*a[i]<=m) r=mid;
					else l=mid;
				}
				cnt+=n-r;
			}
			else{
				ll l=i,r=n;
				while (l+1!=r){
					int mid=towd(l+r);
					if (a[mid]*a[i]<=m) l=mid;
					else r=mid;
				}
				cnt+=l-i;
			}
		}
		if (cnt<k) L=m;
		else R=m;
	}
	cout<<R<<endl;
	return 0;
}
```