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

#define ll long long
const ll MAXN=200001;
ll n,k;
ll a[MAXN];

ll count(ll x){
	ll ret=0;
	ll ok,ng,mid;
	
	for (int i=0; i<n; i++){
		if (a[i]>=0){
			ok=-1,ng=n;
			while (ng-ok>=2){
				mid=(ok+ng)>>1;
				if (a[i]*a[mid]<=x) ok=mid;
				else ng=mid;
			}
			if (ok>=i) ret+=ok;
			else ret+=ok+1;
		}
		else{
			ng=-1,ok=n;
			while (ok-ng>=2){
				mid=(ok+ng)>>1;
				if (a[i]*a[mid]<=x) ok=mid;
				else ng=mid;
			}
			if (ok<=i) ret+=n-1-ok;
			else ret+=n-ok;
		}
	}
	return (ret)>>1;
}

int main(){
	std::ios::sync_with_stdio(false);
    std::cin.tie(0);

	cin>>n>>k;
	for (int i=0; i<n; i++) cin>>a[i];
	sort(a,a+n);
	ll st=-2e18,ed=2e18;
	while (ed-st>=2){
		ll mid=(st+ed)>>1;
		if (count(mid)<k) st=mid;
		else ed=mid;
	}
	cout<<ed<<endl;
	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 ll long long
const ll MAXN=200001;
ll n,k;
ll a[MAXN];

ll count(ll x){
	ll ret=0;
	ll ok,ng,mid;
	
	for (int i=0; i<n; i++){
		if (a[i]>=0){
			ok=-1,ng=n;
			while (ng-ok>=2){
				mid=(ok+ng)>>1;
				if (a[i]*a[mid]<=x) ok=mid;
				else ng=mid;
			}
			if (ok>=i) ret+=ok;
			else ret+=ok+1;
		}
		else{
			ng=-1,ok=n;
			while (ok-ng>=2){
				mid=(ok+ng)>>1;
				if (a[i]*a[mid]<=x) ok=mid;
				else ng=mid;
			}
			if (ok<=i) ret+=n-1-ok;
			else ret+=n-ok;
		}
	}
	return (ret)>>1;
}

int main(){
	std::ios::sync_with_stdio(false);
    std::cin.tie(0);

	cin>>n>>k;
	for (int i=0; i<n; i++) cin>>a[i];
	sort(a,a+n);
	ll st=-2e18,ed=2e18;
	while (ed-st>=2){
		ll mid=(st+ed)>>1;
		if (count(mid)<k) st=mid;
		else ed=mid;
	}
	cout<<ed<<endl;
	return 0;
}
```