#include <bits/stdc++.h>
using namespace std;

#define ll long long

ll a[200001];

int main(){
	std::ios::sync_with_stdio(false);
    std::cin.tie(0);

	ll n;cin>>n;
	ll k;cin>>k;
	for (ll i=0; i<n; i++) cin>>a[i];sort(a,a+n);
	
	ll lo=-2e18,hi=2e18,ans=0;
	while (lo<=hi){
		ll mid=(lo+hi)>>1,cnt=0;
		for (int i=0; i<n; i++){
			if (a[i]<0){
				ll Lo=0,Hi=n-1,Ans=n;
				while (Lo<=Hi){
					ll Mid=(Lo+Hi)>>1;
					if (a[Mid]*a[i]<=mid) Hi=Mid-1,Ans=Mid;
					else Lo=Mid+1;
				}
				cnt+=n-Ans;
			}
			else{
				ll Lo=0,Hi=n-1,Ans=-1;
				while(Lo<=Hi){
					ll Mid=(Lo+Hi)>>1;
					if(a[Mid]*a[i]<=mid) Lo=Mid+1,Ans=Mid;
					else Hi=Mid-1;
				}
				cnt+=Ans+1;
			}
			if (a[i]*a[i]<=mid) cnt--;
		}
		cnt/=2;
		if(cnt>=k) ans=mid,hi=mid-1;
		else lo=mid+1;
	}
	cout<<ans<<endl;
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

ll a[200001];

int main(){
	std::ios::sync_with_stdio(false);
    std::cin.tie(0);

	ll n;cin>>n;
	ll k;cin>>k;
	for (ll i=0; i<n; i++) cin>>a[i];sort(a,a+n);
	
	ll lo=-2e18,hi=2e18,ans=0;
	while (lo<=hi){
		ll mid=(lo+hi)>>1,cnt=0;
		for (int i=0; i<n; i++){
			if (a[i]<0){
				ll Lo=0,Hi=n-1,Ans=n;
				while (Lo<=Hi){
					ll Mid=(Lo+Hi)>>1;
					if (a[Mid]*a[i]<=mid) Hi=Mid-1,Ans=Mid;
					else Lo=Mid+1;
				}
				cnt+=n-Ans;
			}
			else{
				ll Lo=0,Hi=n-1,Ans=-1;
				while(Lo<=Hi){
					ll Mid=(Lo+Hi)>>1;
					if(a[Mid]*a[i]<=mid) Lo=Mid+1,Ans=Mid;
					else Hi=Mid-1;
				}
				cnt+=Ans+1;
			}
			if (a[i]*a[i]<=mid) cnt--;
		}
		cnt/=2;
		if(cnt>=k) ans=mid,hi=mid-1;
		else lo=mid+1;
	}
	cout<<ans<<endl;
	return 0;
}
```