#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
 
const int N=2e5+7;
ll n,k,a[N],b[N],c[N],po,ne;
 
bool check(ll x) {
	int p=ne;ll ret=0;
	for (int i=1;i<=ne;i++) {
		while (p>=1&&b[i]*b[p]<x) p--;
		ret+=p;
	}
	p=1;
	for (int i=po;i>=1;i--) {
		while (p<=po&&c[i]*c[p]<x) p++;
		ret+=po-p+1;
	}
	p=po;
	for (int i=1;i<=n;i++) if (a[i]*a[i]>=x) ret--;
	ret/=2;
	for (int i=ne;i>=1;i--) {
		while (p>=1&&b[i]*c[p]<x) p--;
		ret+=p;
	}
	return n*(n-1)/2-ret>=k;
}
 
int main()
{
	scanf("%lld%lld",&n,&k);
	for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	for (int i=1;i<=n;i++) if (a[i]<0) b[++ne]=a[i];else c[++po]=a[i];
	ll l=-1e18,r=1e18;
	while (l<r) {
		ll mid=(l+r) >> 1;
		if (check(mid)) r=mid;
		else l=mid+1;
	}
	printf("%lld\n",l-1);
	return 0;
}
//老天，AC吧，再不AC，要被打脸的~~(>_<)~~
### 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;

typedef long long ll;
 
const int N=2e5+7;
ll n,k,a[N],b[N],c[N],po,ne;
 
bool check(ll x) {
	int p=ne;ll ret=0;
	for (int i=1;i<=ne;i++) {
		while (p>=1&&b[i]*b[p]<x) p--;
		ret+=p;
	}
	p=1;
	for (int i=po;i>=1;i--) {
		while (p<=po&&c[i]*c[p]<x) p++;
		ret+=po-p+1;
	}
	p=po;
	for (int i=1;i<=n;i++) if (a[i]*a[i]>=x) ret--;
	ret/=2;
	for (int i=ne;i>=1;i--) {
		while (p>=1&&b[i]*c[p]<x) p--;
		ret+=p;
	}
	return n*(n-1)/2-ret>=k;
}
 
int main()
{
	scanf("%lld%lld",&n,&k);
	for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	for (int i=1;i<=n;i++) if (a[i]<0) b[++ne]=a[i];else c[++po]=a[i];
	ll l=-1e18,r=1e18;
	while (l<r) {
		ll mid=(l+r) >> 1;
		if (check(mid)) r=mid;
		else l=mid+1;
	}
	printf("%lld\n",l-1);
	return 0;
}
//老天，AC吧，再不AC，要被打脸的~~(>_<)~~
```