#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef long double ld;
vector<ld> a;
int main(){
	ll i,n,k;
	cin >> n >> k;
	for(i=0;i<n;i++){
		ll x; cin >> x;
		a.push_back(x);
	}
	sort(a.begin(),a.end());
	ll r = 1000000000000000001;
	ll l = -1000000000000000001;
	while(r - l>1){
		ll mid = (l + r)/2,cnt = 0;
		for(i=0;i<n;i++){
			if(a[i]==0){
				if(mid>=0){
					cnt += n - 1;
				}
			}
			if(a[i]>0){
				auto it = upper_bound(a.begin(),a.end(),(ld)mid/a[i]);
				cnt += (it - a.begin());
				if(a[i]*a[i]<=mid) cnt--;
			}
			if(a[i]<0){
				auto it = lower_bound(a.begin(),a.end(),(ld)mid/a[i]);
				cnt += n - (it - a.begin());
				if(a[i]*a[i]<=mid) cnt--;
			}
		}
		cnt /= 2;
		if(cnt>=k) r = mid;
		else l = mid;
	}
	cout << r << endl;
}
			
	
### Prompt

Create a solution in CPP for 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 <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef long double ld;
vector<ld> a;
int main(){
	ll i,n,k;
	cin >> n >> k;
	for(i=0;i<n;i++){
		ll x; cin >> x;
		a.push_back(x);
	}
	sort(a.begin(),a.end());
	ll r = 1000000000000000001;
	ll l = -1000000000000000001;
	while(r - l>1){
		ll mid = (l + r)/2,cnt = 0;
		for(i=0;i<n;i++){
			if(a[i]==0){
				if(mid>=0){
					cnt += n - 1;
				}
			}
			if(a[i]>0){
				auto it = upper_bound(a.begin(),a.end(),(ld)mid/a[i]);
				cnt += (it - a.begin());
				if(a[i]*a[i]<=mid) cnt--;
			}
			if(a[i]<0){
				auto it = lower_bound(a.begin(),a.end(),(ld)mid/a[i]);
				cnt += n - (it - a.begin());
				if(a[i]*a[i]<=mid) cnt--;
			}
		}
		cnt /= 2;
		if(cnt>=k) r = mid;
		else l = mid;
	}
	cout << r << endl;
}
			
	
```