#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
const ll inf=1e18+10;
ll n,k,ans,a[200005];
ll find1(int now,ll x){
	int l=now+1,r=n,best=now;
	while(l<=r){
		int mid=l+r>>1;
		if(a[now]*a[mid]<=x){
			l=mid+1;
			best=mid;
		}
		else r=mid-1;
	}
	return best;
}
ll find2(int now,ll x){
	int l=now+1,r=n,best=n+1;
	while(l<=r){
		int mid=l+r>>1;
		if(a[now]*a[mid]<=x){
			r=mid-1;
			best=mid;
		}
		else l=mid+1;
	}
	return best;
}
bool check(ll x){
	ll cnt=0;   //找比 x 小的积 
	for(int i=1;i<n;i++){
		if(a[i]>=0) cnt+=find1(i,x)-i;
		else cnt+=n-find2(i,x)+1;
	}
	return cnt>=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);
	ll l=-inf,r=inf;
	while(l<=r){
		ll mid=l+r>>1;
		if(check(mid)){
			ans=mid;
			r=mid-1;
		}
		else l=mid+1;
	}
	printf("%lld",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 <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
const ll inf=1e18+10;
ll n,k,ans,a[200005];
ll find1(int now,ll x){
	int l=now+1,r=n,best=now;
	while(l<=r){
		int mid=l+r>>1;
		if(a[now]*a[mid]<=x){
			l=mid+1;
			best=mid;
		}
		else r=mid-1;
	}
	return best;
}
ll find2(int now,ll x){
	int l=now+1,r=n,best=n+1;
	while(l<=r){
		int mid=l+r>>1;
		if(a[now]*a[mid]<=x){
			r=mid-1;
			best=mid;
		}
		else l=mid+1;
	}
	return best;
}
bool check(ll x){
	ll cnt=0;   //找比 x 小的积 
	for(int i=1;i<n;i++){
		if(a[i]>=0) cnt+=find1(i,x)-i;
		else cnt+=n-find2(i,x)+1;
	}
	return cnt>=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);
	ll l=-inf,r=inf;
	while(l<=r){
		ll mid=l+r>>1;
		if(check(mid)){
			ans=mid;
			r=mid-1;
		}
		else l=mid+1;
	}
	printf("%lld",ans);
	return 0;
}
```