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

typedef long long ll;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	ll n, k;
	cin >> n >> k;
	k--;
	vector<ll> arr(n);
	for(ll& x : arr) cin >> x;
	sort(arr.begin(), arr.end());
	
	ll lo = -2e18, hi = 2e18;
	
	while(lo < hi) {
		ll mid = lo + (hi-lo)/2;
		
		ll cnt = 0; // number of products >= mid
		
		for(int i = 0; i < n; i++) {
			if(arr[i] > 0) {
				cnt += lower_bound(arr.begin(), arr.end(), arr[i], [&](ll x, ll y) { return x*y <= mid; }) - arr.begin();
			} else if(arr[i] < 0) {
				cnt += arr.end() - lower_bound(arr.begin(), arr.end(), arr[i], [&](ll x, ll y) { return x*y > mid; });
			} else {
				cnt += mid >= 0 ? n : 0;
			}
			if(arr[i]*arr[i] <= mid) cnt--;
		}
		
		cnt /= 2;
		if(cnt <= k) {
			lo = mid + 1;
		} else {
			hi = mid;
		}
	}
	
	cout << hi << endl;
}

### 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;

typedef long long ll;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	ll n, k;
	cin >> n >> k;
	k--;
	vector<ll> arr(n);
	for(ll& x : arr) cin >> x;
	sort(arr.begin(), arr.end());
	
	ll lo = -2e18, hi = 2e18;
	
	while(lo < hi) {
		ll mid = lo + (hi-lo)/2;
		
		ll cnt = 0; // number of products >= mid
		
		for(int i = 0; i < n; i++) {
			if(arr[i] > 0) {
				cnt += lower_bound(arr.begin(), arr.end(), arr[i], [&](ll x, ll y) { return x*y <= mid; }) - arr.begin();
			} else if(arr[i] < 0) {
				cnt += arr.end() - lower_bound(arr.begin(), arr.end(), arr[i], [&](ll x, ll y) { return x*y > mid; });
			} else {
				cnt += mid >= 0 ? n : 0;
			}
			if(arr[i]*arr[i] <= mid) cnt--;
		}
		
		cnt /= 2;
		if(cnt <= k) {
			lo = mid + 1;
		} else {
			hi = mid;
		}
	}
	
	cout << hi << endl;
}

```