#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
ll k,arr[202020],st,en;
int main()
{
	scanf("%d%lld",&n,&k);
	for(int i=0;i<n;i++) scanf("%lld",arr+i);
	sort(arr,arr+n); st=-2e18; en=2e18;
	while(st+1!=en)
	{
		ll mid=(st+en)/2;
		ll gesu=0;
		for(int i=0;i<n;i++)
		{
			if(arr[i]<0)
			{
				int l=i,r=n;
				while(l+1!=r)
				{
					int m=(l+r)/2;
					if(arr[i]*arr[m]<=mid) r=m;
					else l=m;
				} gesu+=n-r;
			} else
			{
				int l=i,r=n;
				while(l+1!=r)
				{
					int m=(l+r)/2;
					if(arr[i]*arr[m]<=mid) l=m;
					else r=m;
				} gesu+=l-i;
			} 
		} if(gesu<k) st=mid;
		else en=mid;
	} printf("%lld",en);
}



### Prompt

Please formulate 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;
typedef long long ll;
int n;
ll k,arr[202020],st,en;
int main()
{
	scanf("%d%lld",&n,&k);
	for(int i=0;i<n;i++) scanf("%lld",arr+i);
	sort(arr,arr+n); st=-2e18; en=2e18;
	while(st+1!=en)
	{
		ll mid=(st+en)/2;
		ll gesu=0;
		for(int i=0;i<n;i++)
		{
			if(arr[i]<0)
			{
				int l=i,r=n;
				while(l+1!=r)
				{
					int m=(l+r)/2;
					if(arr[i]*arr[m]<=mid) r=m;
					else l=m;
				} gesu+=n-r;
			} else
			{
				int l=i,r=n;
				while(l+1!=r)
				{
					int m=(l+r)/2;
					if(arr[i]*arr[m]<=mid) l=m;
					else r=m;
				} gesu+=l-i;
			} 
		} if(gesu<k) st=mid;
		else en=mid;
	} printf("%lld",en);
}



```