#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(x) x.begin(), x.end()
#define fi first
#define se second
#define mp make_pair
#define pb push_back
const ll INF = 1e18;

ll intfloor(ll a, ll b){
	if(a < 0) return (a-b+1)/b;
	else return a/b;
}

ll intceil(ll a, ll b){
	if(a > 0) return (a+b-1)/b;
	else return a/b;
}

int main(){
	ll n, k;
	cin >> n >> k;
	vector<int> A(n);
	for(int i = 0; i < n ; i++){
		cin >> A[i];
	}
	sort(all(A));
	ll ok = INF, ng = -INF - 1;
	while(ok - ng > 1){
		ll mid = (ok + ng) / 2;
		ll num = 0;
		for(int i = 0; i < n; i++){
			if(A[i] == 0){
				if(mid >= 0) num += i;
			}
			else if(A[i] > 0){
				num += upper_bound(A.begin(), A.begin() + i, intfloor(mid, A[i])) - A.begin();
			}
			else{
				num += A.begin() + i -lower_bound(A.begin(), A.begin() + i, intceil(-mid, -A[i]));
			}
		}
		if(num >= k) ok = mid;
		else ng = mid;
	}
	cout << ok << endl;
	return 0;
}

### 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;
#define all(x) x.begin(), x.end()
#define fi first
#define se second
#define mp make_pair
#define pb push_back
const ll INF = 1e18;

ll intfloor(ll a, ll b){
	if(a < 0) return (a-b+1)/b;
	else return a/b;
}

ll intceil(ll a, ll b){
	if(a > 0) return (a+b-1)/b;
	else return a/b;
}

int main(){
	ll n, k;
	cin >> n >> k;
	vector<int> A(n);
	for(int i = 0; i < n ; i++){
		cin >> A[i];
	}
	sort(all(A));
	ll ok = INF, ng = -INF - 1;
	while(ok - ng > 1){
		ll mid = (ok + ng) / 2;
		ll num = 0;
		for(int i = 0; i < n; i++){
			if(A[i] == 0){
				if(mid >= 0) num += i;
			}
			else if(A[i] > 0){
				num += upper_bound(A.begin(), A.begin() + i, intfloor(mid, A[i])) - A.begin();
			}
			else{
				num += A.begin() + i -lower_bound(A.begin(), A.begin() + i, intceil(-mid, -A[i]));
			}
		}
		if(num >= k) ok = mid;
		else ng = mid;
	}
	cout << ok << endl;
	return 0;
}

```