#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=200005;
int n,a[N];
LL k;
bool check(LL x){
	LL cnt=0;
	for(int i=1;i<=n;i++){
		if(a[i]<=0){
			int l=0,r=n,mid;
			while(l<r-1){
				mid=(l+r)>>1;
				if(1LL*a[i]*a[mid]<=x)r=mid;
				else l=mid;
			}
			if(r<=i)cnt+=n-r;
			else cnt+=n-r+1;
			if(1LL*a[i]*a[r]>x)cnt--;
		}else{
			int l=1,r=n+1,mid;
			while(l<r-1){
				mid=(l+r)>>1;
				if(1LL*a[i]*a[mid]<=x)l=mid;
				else r=mid;
			}
			if(l<=i)cnt+=l;
			else cnt+=l-1;
			if(1LL*a[i]*a[l]>x)cnt--;
		}
		if(cnt>=k*2)return 1;
	}
	return 0;
}
int main(){
	LL l=-1e18-1,r=1e18,mid;
	scanf("%d%lld",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	sort(a+1,a+1+n);	
	//printf("%d\n",check(-12));
	while(l<r-1){
		mid=(l+r)>>1;
		if(check(mid))r=mid;
		else l=mid;
	}
	printf("%lld\n",r);
	return 0;
}
### Prompt

Please create a solution in cpp 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;
typedef long long LL;
const int N=200005;
int n,a[N];
LL k;
bool check(LL x){
	LL cnt=0;
	for(int i=1;i<=n;i++){
		if(a[i]<=0){
			int l=0,r=n,mid;
			while(l<r-1){
				mid=(l+r)>>1;
				if(1LL*a[i]*a[mid]<=x)r=mid;
				else l=mid;
			}
			if(r<=i)cnt+=n-r;
			else cnt+=n-r+1;
			if(1LL*a[i]*a[r]>x)cnt--;
		}else{
			int l=1,r=n+1,mid;
			while(l<r-1){
				mid=(l+r)>>1;
				if(1LL*a[i]*a[mid]<=x)l=mid;
				else r=mid;
			}
			if(l<=i)cnt+=l;
			else cnt+=l-1;
			if(1LL*a[i]*a[l]>x)cnt--;
		}
		if(cnt>=k*2)return 1;
	}
	return 0;
}
int main(){
	LL l=-1e18-1,r=1e18,mid;
	scanf("%d%lld",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	sort(a+1,a+1+n);	
	//printf("%d\n",check(-12));
	while(l<r-1){
		mid=(l+r)>>1;
		if(check(mid))r=mid;
		else l=mid;
	}
	printf("%lld\n",r);
	return 0;
}
```