#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=2e5+10;
ll n,k,l,r,a[N];
void pt(ll a,ll b,ll c)
{
	printf("a[%d]:%lld num:%lld\n",a,b,c);
}
ll cal(ll x)
{
	ll num=0;
	if(x>=0)
	{
		for(int i=0;i<n;++i)
		{
			if(a[i]==0)num+=n;
			else if(a[i]>0)num+=upper_bound(a,a+n,x/a[i])-a;
			else num+=n-(lower_bound(a,a+n,x/a[i])-a);
		}
	}
	else
	{
		for(int i=0;i<n;++i)
		{
			if(a[i]==0)continue;
			if(a[i]>0)num+=upper_bound(a,a+n,(x-a[i]+1)/a[i])-a;
			else num+=n-(lower_bound(a,a+n,(x+a[i]+1)/a[i])-a);
		}
	}
	for(int i=0;i<n;++i)
		if(a[i]*a[i]<=x)num--;
	num/=2;
	return num;
}
int main()
{
	scanf("%lld%lld",&n,&k);
	for(int i=0;i<n;++i)
	scanf("%lld",&a[i]);
	sort(a,a+n);
	l=-1e18,r=1e18;
	while(l<=r)
	{
		ll mid=(l+r)/2;
		if(cal(mid)>=k)r=mid-1;
		else l=mid+1;
	}
	printf("%lld\n",l);
	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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=2e5+10;
ll n,k,l,r,a[N];
void pt(ll a,ll b,ll c)
{
	printf("a[%d]:%lld num:%lld\n",a,b,c);
}
ll cal(ll x)
{
	ll num=0;
	if(x>=0)
	{
		for(int i=0;i<n;++i)
		{
			if(a[i]==0)num+=n;
			else if(a[i]>0)num+=upper_bound(a,a+n,x/a[i])-a;
			else num+=n-(lower_bound(a,a+n,x/a[i])-a);
		}
	}
	else
	{
		for(int i=0;i<n;++i)
		{
			if(a[i]==0)continue;
			if(a[i]>0)num+=upper_bound(a,a+n,(x-a[i]+1)/a[i])-a;
			else num+=n-(lower_bound(a,a+n,(x+a[i]+1)/a[i])-a);
		}
	}
	for(int i=0;i<n;++i)
		if(a[i]*a[i]<=x)num--;
	num/=2;
	return num;
}
int main()
{
	scanf("%lld%lld",&n,&k);
	for(int i=0;i<n;++i)
	scanf("%lld",&a[i]);
	sort(a,a+n);
	l=-1e18,r=1e18;
	while(l<=r)
	{
		ll mid=(l+r)/2;
		if(cal(mid)>=k)r=mid-1;
		else l=mid+1;
	}
	printf("%lld\n",l);
	return 0;
}
```