#include<bits/stdc++.h>
using namespace std;
#define int long long
int n,k,num[200100];
bool che(int ip){
	int res=0;
	for(int i=1;i<=n;i++){
		if(num[i]==0)res+=(n-1)*(ip>=0);
		if(num[i]>0){
			if(num[i]*num[i]<=ip)res--;
			if(num[i]*num[1]>ip)continue;
			int l=1,r=n;
			while(l<r){
				int mid=(l+r+1)>>1;
				if(num[i]*num[mid]<=ip)l=mid;
				else r=mid-1;
			}
//			printf("%lld:>:%lld\n",i,l);
			res+=l;
		}
		if(num[i]<0){
			if(num[i]*num[i]<=ip)res--;
			if(num[i]*num[n]>ip)continue;
			int l=1,r=n;
			while(l<r){
				int mid=(l+r)>>1;
				if(num[i]*num[mid]<=ip)r=mid;
				else l=mid+1;
			}
//			printf("%lld:<:%lld\n",i,r);
			res+=n-r+1;
		}
	}
//	printf("%lld:%lld\n",ip,res);
	res>>=1;
	return res>=k;
}
signed main(){
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)scanf("%lld",&num[i]);
	int l=-1e18,r=1e18;
	sort(num+1,num+n+1);
	while(l<r){
		int mid=(l+r)>>1;
		if(che(mid))r=mid;
		else l=mid+1;
	}
	printf("%lld\n",r);
	return 0;
}
### Prompt

Your task is to create 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 int long long
int n,k,num[200100];
bool che(int ip){
	int res=0;
	for(int i=1;i<=n;i++){
		if(num[i]==0)res+=(n-1)*(ip>=0);
		if(num[i]>0){
			if(num[i]*num[i]<=ip)res--;
			if(num[i]*num[1]>ip)continue;
			int l=1,r=n;
			while(l<r){
				int mid=(l+r+1)>>1;
				if(num[i]*num[mid]<=ip)l=mid;
				else r=mid-1;
			}
//			printf("%lld:>:%lld\n",i,l);
			res+=l;
		}
		if(num[i]<0){
			if(num[i]*num[i]<=ip)res--;
			if(num[i]*num[n]>ip)continue;
			int l=1,r=n;
			while(l<r){
				int mid=(l+r)>>1;
				if(num[i]*num[mid]<=ip)r=mid;
				else l=mid+1;
			}
//			printf("%lld:<:%lld\n",i,r);
			res+=n-r+1;
		}
	}
//	printf("%lld:%lld\n",ip,res);
	res>>=1;
	return res>=k;
}
signed main(){
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)scanf("%lld",&num[i]);
	int l=-1e18,r=1e18;
	sort(num+1,num+n+1);
	while(l<r){
		int mid=(l+r)>>1;
		if(che(mid))r=mid;
		else l=mid+1;
	}
	printf("%lld\n",r);
	return 0;
}
```