#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=200001;
const ll inf=(ll)1e18;
ll n,a[N],k,num=0;

int Find(int now,ll x)
{
	int l=now+1,r=n,best=now;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(a[now]*a[mid]<=x) l=mid+1,best=mid;
		else r=mid-1;
	}
	return best;
}

int Find2(int now,ll x)
{
	int l=now+1,r=n,best=n+1;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(a[now]*a[mid]<=x) r=mid-1,best=mid;
		else l=mid+1;
	}
	return best;
}

bool check(ll x)
{
	ll ret=0;
	for(int i=1;i<n;i++)
	{
		if(a[i]>=0) ret+=Find(i,x)-i;
		else ret+=n-Find2(i,x)+1;
		if(x==-12) cout<<ret<<endl;
	}
	return ret>=k;
}

int main()
{
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	ll l=-inf,r=inf,ans=-1;
	while(l<=r)
	{
		ll mid=(l+r)>>1;
		if(check(mid)) r=mid-1,ans=mid;
		else l=mid+1;
	}
	cout<<ans<<endl;
	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;
typedef long long ll;
const int N=200001;
const ll inf=(ll)1e18;
ll n,a[N],k,num=0;

int Find(int now,ll x)
{
	int l=now+1,r=n,best=now;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(a[now]*a[mid]<=x) l=mid+1,best=mid;
		else r=mid-1;
	}
	return best;
}

int Find2(int now,ll x)
{
	int l=now+1,r=n,best=n+1;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(a[now]*a[mid]<=x) r=mid-1,best=mid;
		else l=mid+1;
	}
	return best;
}

bool check(ll x)
{
	ll ret=0;
	for(int i=1;i<n;i++)
	{
		if(a[i]>=0) ret+=Find(i,x)-i;
		else ret+=n-Find2(i,x)+1;
		if(x==-12) cout<<ret<<endl;
	}
	return ret>=k;
}

int main()
{
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	ll l=-inf,r=inf,ans=-1;
	while(l<=r)
	{
		ll mid=(l+r)>>1;
		if(check(mid)) r=mid-1,ans=mid;
		else l=mid+1;
	}
	cout<<ans<<endl;
	return 0;
}
```