#include <bits/stdc++.h>
using namespace std;
#define N 200005
#define ll long long
int n;ll m,L,R,ans,a[N];
bool chk(ll x)
{
	ll cnt=0,t1,t2;
	for(int i=1;i<=n;++i)
		if(a[i])
		{
			t1=t2=x/a[i];
			if(x%a[i])
			{
				if((x<0 && a[i]>0) || (x>0 && a[i]<0)) --t1;
				else ++t2;
			}
			if(a[i]<0) cnt+=n+1-(upper_bound(a+1,a+n+1,t1)-a);
			if(a[i]>0) cnt+=lower_bound(a+1,a+n+1,t2)-a-1;
			if(a[i]*a[i]<x) --cnt;
		}else if(x>0) cnt+=n-1;cnt>>=1;return cnt<m;
}
int main()
{
	scanf("%d %lld",&n,&m);L=-1e18;R=1e18;
	for(int i=1;i<=n;++i) scanf("%lld",&a[i]);sort(a+1,a+n+1);
	while(L<=R) {ll mid=L+R>>1;if(chk(mid)) L=mid+1,ans=mid;else R=mid-1;}
	printf("%lld\n",ans);return 0;
}
### Prompt

Your task is to create 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;
#define N 200005
#define ll long long
int n;ll m,L,R,ans,a[N];
bool chk(ll x)
{
	ll cnt=0,t1,t2;
	for(int i=1;i<=n;++i)
		if(a[i])
		{
			t1=t2=x/a[i];
			if(x%a[i])
			{
				if((x<0 && a[i]>0) || (x>0 && a[i]<0)) --t1;
				else ++t2;
			}
			if(a[i]<0) cnt+=n+1-(upper_bound(a+1,a+n+1,t1)-a);
			if(a[i]>0) cnt+=lower_bound(a+1,a+n+1,t2)-a-1;
			if(a[i]*a[i]<x) --cnt;
		}else if(x>0) cnt+=n-1;cnt>>=1;return cnt<m;
}
int main()
{
	scanf("%d %lld",&n,&m);L=-1e18;R=1e18;
	for(int i=1;i<=n;++i) scanf("%lld",&a[i]);sort(a+1,a+n+1);
	while(L<=R) {ll mid=L+R>>1;if(chk(mid)) L=mid+1,ans=mid;else R=mid-1;}
	printf("%lld\n",ans);return 0;
}
```