#include<bits/stdc++.h>
using namespace std;
int n,a[200005],z[200005],f[200005],zt,ft,zr;long long k,l,r;
inline int chck(long long v)
{
	long long cnt=0;
	if(v>=0)
	{
		cnt=1ll*zr*(zt+ft)+1ll*zt*ft+1ll*zr*(zr-1)/2;
		for(int i=1;i<=zt;i++) cnt+=upper_bound(z+i+1,z+zt+1,v/z[i])-z-1-i;
		for(int i=1;i<=ft;i++) cnt+=upper_bound(f+i+1,f+ft+1,v/f[i])-f-1-i;
	}
	else
	{
		v=-v;
		for(int i=1;i<=ft;i++) cnt+=z+zt+1-lower_bound(z+1,z+zt+1,v/f[i]+!!(v%f[i]));
	}
	return cnt>=k;
}
int main()
{
	scanf("%d%lld",&n,&k),zt=ft=zr=0,r=(long long)1e18,l=-r;
	for(int i=1;i<=n;i++) scanf("%d",a+i);
	for(int i=1;i<=n;i++) if(a[i]>0) z[++zt]=a[i];else if(a[i]<0) f[++ft]=-a[i];else zr++;
	sort(z+1,z+zt+1),sort(f+1,f+ft+1);
	while(l<r) {if(chck((l+r)>>1)) r=(l+r)>>1;else l=((l+r)>>1)+1;}
	return printf("%lld\n",l),0;
}
### Prompt

Develop a solution in Cpp 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<bits/stdc++.h>
using namespace std;
int n,a[200005],z[200005],f[200005],zt,ft,zr;long long k,l,r;
inline int chck(long long v)
{
	long long cnt=0;
	if(v>=0)
	{
		cnt=1ll*zr*(zt+ft)+1ll*zt*ft+1ll*zr*(zr-1)/2;
		for(int i=1;i<=zt;i++) cnt+=upper_bound(z+i+1,z+zt+1,v/z[i])-z-1-i;
		for(int i=1;i<=ft;i++) cnt+=upper_bound(f+i+1,f+ft+1,v/f[i])-f-1-i;
	}
	else
	{
		v=-v;
		for(int i=1;i<=ft;i++) cnt+=z+zt+1-lower_bound(z+1,z+zt+1,v/f[i]+!!(v%f[i]));
	}
	return cnt>=k;
}
int main()
{
	scanf("%d%lld",&n,&k),zt=ft=zr=0,r=(long long)1e18,l=-r;
	for(int i=1;i<=n;i++) scanf("%d",a+i);
	for(int i=1;i<=n;i++) if(a[i]>0) z[++zt]=a[i];else if(a[i]<0) f[++ft]=-a[i];else zr++;
	sort(z+1,z+zt+1),sort(f+1,f+ft+1);
	while(l<r) {if(chck((l+r)>>1)) r=(l+r)>>1;else l=((l+r)>>1)+1;}
	return printf("%lld\n",l),0;
}
```