#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll mod = 1e9 +7;
const ll MAXN = 1e6 + 5;

ll n,k;
ll a[MAXN];
//判断在当前的数之前有多少个数
ll check(ll x){
	ll res=0;
	for(int i=0;i<n;i++){
		if(a[i]<0){
			ll L=i,R=n;
			while(R-L>1){
				ll mid=L+R>>1;
				if(a[i]*a[mid]<=x){//这里要注意和下面不一样的是 当a[i]<0的时候mid取小 乘积反而会增大
					R=mid;
				}else{
					L=mid;
				}
			}
			res+=n-R;
		}else{
			ll L=i,R=n;
			while(R-L>1){
				ll mid=L+R>>1;
				if(a[i]*a[mid]<=x){//a[i]>0 mid取大 乘积增大
					L=mid;
				}else{
					R=mid;
				}
			}
			res+=L-i;
		}
	}
	return res;
}


int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a,a+n);
	ll L=-1e18,R=1e18;
	while(R-L>1){
		ll Mid=L+R>>1;
		if(check(Mid)<k){
			L=Mid;
		}else{
			R=Mid;
		}
	}
	cout<<R;
	return 0;
}


### Prompt

Please provide a Cpp coded solution to the problem described below:

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 INF = 0x3f3f3f3f;
const ll mod = 1e9 +7;
const ll MAXN = 1e6 + 5;

ll n,k;
ll a[MAXN];
//判断在当前的数之前有多少个数
ll check(ll x){
	ll res=0;
	for(int i=0;i<n;i++){
		if(a[i]<0){
			ll L=i,R=n;
			while(R-L>1){
				ll mid=L+R>>1;
				if(a[i]*a[mid]<=x){//这里要注意和下面不一样的是 当a[i]<0的时候mid取小 乘积反而会增大
					R=mid;
				}else{
					L=mid;
				}
			}
			res+=n-R;
		}else{
			ll L=i,R=n;
			while(R-L>1){
				ll mid=L+R>>1;
				if(a[i]*a[mid]<=x){//a[i]>0 mid取大 乘积增大
					L=mid;
				}else{
					R=mid;
				}
			}
			res+=L-i;
		}
	}
	return res;
}


int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>k;
	for(int i=0;i<n;i++) cin>>a[i];
	sort(a,a+n);
	ll L=-1e18,R=1e18;
	while(R-L>1){
		ll Mid=L+R>>1;
		if(check(Mid)<k){
			L=Mid;
		}else{
			R=Mid;
		}
	}
	cout<<R;
	return 0;
}


```