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

typedef long long ll;
int n;
ll k,a[202020];
ll st=-2e18,en=2e18;

int main() {
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a,a+n);
	while(st+1!=en) {
		ll mid=(st+en)/2,cnt=0;
		for(int i=0;i<n;i++) {
			if(a[i]<0) {
				int l=i,r=n;
				while(l+1!=r) {
					int m=(l+r)/2;
					if(a[i]*a[m]<=mid) r=m;
					else l=m;
				}
				cnt+=n-r;
			} else {
				int l=i,r=n;
				while(l+1!=r) {
					int m=(l+r)/2;
					if(a[i]*a[m]<=mid) l=m;
					else r=m;
				}
				cnt+=l-i;
			}
		}
		
		if(cnt<k) st=mid;
		else en=mid;
	}
	cout<<en<<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;

typedef long long ll;
int n;
ll k,a[202020];
ll st=-2e18,en=2e18;

int main() {
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a,a+n);
	while(st+1!=en) {
		ll mid=(st+en)/2,cnt=0;
		for(int i=0;i<n;i++) {
			if(a[i]<0) {
				int l=i,r=n;
				while(l+1!=r) {
					int m=(l+r)/2;
					if(a[i]*a[m]<=mid) r=m;
					else l=m;
				}
				cnt+=n-r;
			} else {
				int l=i,r=n;
				while(l+1!=r) {
					int m=(l+r)/2;
					if(a[i]*a[m]<=mid) l=m;
					else r=m;
				}
				cnt+=l-i;
			}
		}
		
		if(cnt<k) st=mid;
		else en=mid;
	}
	cout<<en<<endl;
	return 0;
}
```