#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<functional>
#include<vector>

using namespace std;

typedef long long ll;

ll N, K;
ll a[200000 + 100];

int my_b_search(ll key, ll val, bool mode) {
	//mode -> trueならば昇順での二分探索 falseなら降順での二分探索
	int ok = mode ? -1 : N;
	int ng = mode ? N : -1;
	while (abs(ok - ng) > 1) {
		int mid = (ok + ng) / 2;
		if (val * a[mid] <= key)ok = mid;
		else ng = mid;
	}
	return ok;
}

bool f(ll key) {
	ll cnt = 0;
	for (int i = 0; i < N; i++) {
		if (a[i] >= 0) {
			cnt += my_b_search(key, a[i], true) + 1;
		}
		else {
			cnt += N - my_b_search(key, a[i], false);
		}
		if (key >= a[i] * a[i])cnt--;
	}
	return cnt >= 2 * K;
}

int main() {
	cin >> N >> K;
	for (int i = 0; i < N; i++)cin >> a[i];

	sort(a, a + N);
	
	ll ng = -1e18 - 100, ok = 1e18 + 100;
	while (abs(ok - ng) > 1) {
		ll mid = (ok + ng) / 2;
		if (f(mid))ok = mid;
		else ng = mid;
	}
	cout << ok << endl;
	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<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<functional>
#include<vector>

using namespace std;

typedef long long ll;

ll N, K;
ll a[200000 + 100];

int my_b_search(ll key, ll val, bool mode) {
	//mode -> trueならば昇順での二分探索 falseなら降順での二分探索
	int ok = mode ? -1 : N;
	int ng = mode ? N : -1;
	while (abs(ok - ng) > 1) {
		int mid = (ok + ng) / 2;
		if (val * a[mid] <= key)ok = mid;
		else ng = mid;
	}
	return ok;
}

bool f(ll key) {
	ll cnt = 0;
	for (int i = 0; i < N; i++) {
		if (a[i] >= 0) {
			cnt += my_b_search(key, a[i], true) + 1;
		}
		else {
			cnt += N - my_b_search(key, a[i], false);
		}
		if (key >= a[i] * a[i])cnt--;
	}
	return cnt >= 2 * K;
}

int main() {
	cin >> N >> K;
	for (int i = 0; i < N; i++)cin >> a[i];

	sort(a, a + N);
	
	ll ng = -1e18 - 100, ok = 1e18 + 100;
	while (abs(ok - ng) > 1) {
		ll mid = (ok + ng) / 2;
		if (f(mid))ok = mid;
		else ng = mid;
	}
	cout << ok << endl;
	return 0;
}

```