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

vector<int> A;
int N,K;
const int inf=1000000000000000001;

void solve(int &l,int &r){
	//l,r以下の数が何個あるかで二分探索
	while(l+1<r){
		int x=(l+r)/2;
		int sum=0;
		for(int i=0;i<N;i++){
			if(A[i]>=0){
				int left=-1,right=N;
				while(left+1<right){
					int c=(left+right)/2;
					if(A[c]*A[i]<x)left=c;
					else right=c;
				}
				sum+=right;
			}else{
				int left=-1,right=N;
				while(left+1<right){
					int c=(left+right)/2;
					if(A[c]*A[i]<x)right=c;
					else left=c;
				}
				sum+=N-right;
			}
			if(A[i]*A[i]<x)sum--;
		}
		sum/=2;
		if(sum<K)l=x;
		else r=x;
	}	
}

signed main(){
	cin>>N>>K;
	A.assign(N,0);
	for(int i=0;i<N;i++)cin>>A[i];
	sort(A.begin(),A.end());
	int l=-inf,r=inf;
	solve(l,r);
	cout<<l<<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>
#define int long long
using namespace std;

vector<int> A;
int N,K;
const int inf=1000000000000000001;

void solve(int &l,int &r){
	//l,r以下の数が何個あるかで二分探索
	while(l+1<r){
		int x=(l+r)/2;
		int sum=0;
		for(int i=0;i<N;i++){
			if(A[i]>=0){
				int left=-1,right=N;
				while(left+1<right){
					int c=(left+right)/2;
					if(A[c]*A[i]<x)left=c;
					else right=c;
				}
				sum+=right;
			}else{
				int left=-1,right=N;
				while(left+1<right){
					int c=(left+right)/2;
					if(A[c]*A[i]<x)right=c;
					else left=c;
				}
				sum+=N-right;
			}
			if(A[i]*A[i]<x)sum--;
		}
		sum/=2;
		if(sum<K)l=x;
		else r=x;
	}	
}

signed main(){
	cin>>N>>K;
	A.assign(N,0);
	for(int i=0;i<N;i++)cin>>A[i];
	sort(A.begin(),A.end());
	int l=-inf,r=inf;
	solve(l,r);
	cout<<l<<endl;
	return 0;
}

```