#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
typedef long long LL;
int a[N],b[N];
LL k,s1,s2,s3;
bool check(LL x){
	LL ans=0;
	if(x>=0){
		ans+=s3*(s1+s2)+s3*(s3-1)/2+s1*s2;
		int r=s1-1;
		for(int i=0;i<r;i++){
			while(1ll*a[i]*a[r]>x)r--;
			if(i>=r)break;
			ans+=r-i; 
		}
		r=s2-1;
		for(int i=0;i<r;i++){
			while(1ll*b[i]*b[r]>x)r--;
			if(i>=r)break;
			ans+=r-i;
		}
	}
	else{
		x=-x;
		int l=0;
		for(int i=s2-1;i>=0;i--){
			while(1ll*a[l]*b[i]<x&&l<s1)l++;
			ans+=s1-l;
		}
	}
	return ans>=k;
}
int main(){
	int n,x;LL l=-2e18,r=2e18,mid;scanf("%d%lld",&n,&k);
	for(int i=0;i<n;i++){
		scanf("%d",&x);
		if(x>0)a[s1++]=x;
		else if(x<0)b[s2++]=-x;
		else s3++;
	}
	sort(a,a+s1),sort(b,b+s2);
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid))r=mid-1;
		else l=mid+1;
	}
	printf("%lld",l);
	return 0;
}
### Prompt

In CPP, your task is to solve 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;
const int N=2e5+5;
typedef long long LL;
int a[N],b[N];
LL k,s1,s2,s3;
bool check(LL x){
	LL ans=0;
	if(x>=0){
		ans+=s3*(s1+s2)+s3*(s3-1)/2+s1*s2;
		int r=s1-1;
		for(int i=0;i<r;i++){
			while(1ll*a[i]*a[r]>x)r--;
			if(i>=r)break;
			ans+=r-i; 
		}
		r=s2-1;
		for(int i=0;i<r;i++){
			while(1ll*b[i]*b[r]>x)r--;
			if(i>=r)break;
			ans+=r-i;
		}
	}
	else{
		x=-x;
		int l=0;
		for(int i=s2-1;i>=0;i--){
			while(1ll*a[l]*b[i]<x&&l<s1)l++;
			ans+=s1-l;
		}
	}
	return ans>=k;
}
int main(){
	int n,x;LL l=-2e18,r=2e18,mid;scanf("%d%lld",&n,&k);
	for(int i=0;i<n;i++){
		scanf("%d",&x);
		if(x>0)a[s1++]=x;
		else if(x<0)b[s2++]=-x;
		else s3++;
	}
	sort(a,a+s1),sort(b,b+s2);
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid))r=mid-1;
		else l=mid+1;
	}
	printf("%lld",l);
	return 0;
}
```