#include <bits/stdc++.h>
using namespace std;
int n;
long long a[200005];
long long k,l,r,mid,ans;
long long work(long long a,long long b) {
	return a / b + (((a ^ b) >= 0) and a % b);
}
long long _work(long long a, long long b) {
	return a / b - (((a ^ b) < 0) and a % b);
}
long long check(long long w){
	long long cnt=0,div;
	for (register int i=1;i<=n;++i){
		if (a[i]==0){
			if (w>=0) cnt+=n-1; 
		}
		if (a[i]>0){
		    div=_work(w,a[i]);
			int pos=upper_bound(a+1,a+1+n,div)-a;
			--pos;
			if (pos>=i) cnt+=pos-1;
			else cnt+=pos;
		}
		if (a[i]<0){
			div=work(w,a[i]);
			int pos=lower_bound(a+1,a+1+n,div)-a;
			if (pos<=i) cnt+=n-pos;
			else cnt+=n-pos+1;
		}
	}
	return cnt/2;
}
int main(){
	scanf ("%d%lld",&n,&k);
	for (register int i=1;i<=n;++i) scanf ("%lld",&a[i]);
	sort (a+1,a+1+n);
	l=-1e18;r=1e18;
	while (l<=r){
		mid=l+r>>1;
		if (check(mid)>=k) ans=mid,r=mid-1;
		else l=mid+1;
	}
	printf ("%lld\n",ans);
	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;
int n;
long long a[200005];
long long k,l,r,mid,ans;
long long work(long long a,long long b) {
	return a / b + (((a ^ b) >= 0) and a % b);
}
long long _work(long long a, long long b) {
	return a / b - (((a ^ b) < 0) and a % b);
}
long long check(long long w){
	long long cnt=0,div;
	for (register int i=1;i<=n;++i){
		if (a[i]==0){
			if (w>=0) cnt+=n-1; 
		}
		if (a[i]>0){
		    div=_work(w,a[i]);
			int pos=upper_bound(a+1,a+1+n,div)-a;
			--pos;
			if (pos>=i) cnt+=pos-1;
			else cnt+=pos;
		}
		if (a[i]<0){
			div=work(w,a[i]);
			int pos=lower_bound(a+1,a+1+n,div)-a;
			if (pos<=i) cnt+=n-pos;
			else cnt+=n-pos+1;
		}
	}
	return cnt/2;
}
int main(){
	scanf ("%d%lld",&n,&k);
	for (register int i=1;i<=n;++i) scanf ("%lld",&a[i]);
	sort (a+1,a+1+n);
	l=-1e18;r=1e18;
	while (l<=r){
		mid=l+r>>1;
		if (check(mid)>=k) ans=mid,r=mid-1;
		else l=mid+1;
	}
	printf ("%lld\n",ans);
	return 0;
}
```