#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll n,k;
ll a[200005];
ll l=-1e18-5,r=1e18+5,mid,ans;
int main()
{
//	freopen("data1.in","r",stdin);
//	freopen("jindian.out","w",stdout);
	cin>>n>>k;
	for(int i=1;i<=n;++i)
	{
		cin>>a[i];
	}
	sort(a+1,a+1+n);
//	for(int i=1;i<=n;++i)
//  {
//      printf("%lld ",a[i]);
//  }
//  printf("\n");
	while(l<=r)
	{
		mid=(l+r)>>1;
		ll cnt=0;
//		printf("%lld\n",mid);
		for(int i=1;i<n;++i)
		{
			int li=i+1,ri=n,midi,tmp=0;
			while(li<=ri)
			{
				midi=(li+ri)>>1;
				if(a[i]*a[midi]<=mid)
				{
					if(a[i]>0)
					li=midi+1;
					else
					ri=midi-1;
					tmp=midi;
				}
				else
				{
					if(a[i]>0)
					ri=midi-1;
					else
					li=midi+1;
				}
			}
//			printf("%lld %lld\n",a[i],a[tmp]);
			if(tmp==0)
			continue;
			if(a[i]>0)
			cnt+=tmp-i;
			else
			cnt+=n-tmp+1;
		}
//		printf("%d\n",cnt);
		if(cnt>=k)
		{
			r=mid-1;
			ans=mid;
		}
		else
		{
			l=mid+1;
		}
	}
	printf("%lld\n",ans);
}

### 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;
#define ll long long
ll n,k;
ll a[200005];
ll l=-1e18-5,r=1e18+5,mid,ans;
int main()
{
//	freopen("data1.in","r",stdin);
//	freopen("jindian.out","w",stdout);
	cin>>n>>k;
	for(int i=1;i<=n;++i)
	{
		cin>>a[i];
	}
	sort(a+1,a+1+n);
//	for(int i=1;i<=n;++i)
//  {
//      printf("%lld ",a[i]);
//  }
//  printf("\n");
	while(l<=r)
	{
		mid=(l+r)>>1;
		ll cnt=0;
//		printf("%lld\n",mid);
		for(int i=1;i<n;++i)
		{
			int li=i+1,ri=n,midi,tmp=0;
			while(li<=ri)
			{
				midi=(li+ri)>>1;
				if(a[i]*a[midi]<=mid)
				{
					if(a[i]>0)
					li=midi+1;
					else
					ri=midi-1;
					tmp=midi;
				}
				else
				{
					if(a[i]>0)
					ri=midi-1;
					else
					li=midi+1;
				}
			}
//			printf("%lld %lld\n",a[i],a[tmp]);
			if(tmp==0)
			continue;
			if(a[i]>0)
			cnt+=tmp-i;
			else
			cnt+=n-tmp+1;
		}
//		printf("%d\n",cnt);
		if(cnt>=k)
		{
			r=mid-1;
			ans=mid;
		}
		else
		{
			l=mid+1;
		}
	}
	printf("%lld\n",ans);
}

```