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

using i64 = int64_t;

int main() {
	int N;
	i64 K;
	cin >> N >> K;
	
	vector<i64> A(N);
	for (int i=0; i<N; ++i) cin >> A[i];
	sort(A.begin(), A.end(), greater<i64>());
	
	auto less_count = [&](i64 x) {
		i64 cnt = 0;
		for (int i=1; i<N; ++i) {
			int lb = -1, ub = i;
			while (ub - lb > 1) {
				int med = (lb + ub) / 2;
				if ((A[i] > 0 and A[i] * A[med] > x) or (A[i] <= 0 and A[i] * A[med] <= x)) lb = med;
				else ub = med;
			}
			cnt += (A[i] > 0 ? i - ub : ub);
		}
		return cnt;
	};
		
	i64 lb = -1e18 - 1, ub = 1e18;
	while (ub - lb > 1) {
		i64 med = (lb + ub) / 2;
		if (K <= less_count(med)) ub = med;
		else lb = med;
	}
	cout << ub << endl;
}
### Prompt

Generate 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;

using i64 = int64_t;

int main() {
	int N;
	i64 K;
	cin >> N >> K;
	
	vector<i64> A(N);
	for (int i=0; i<N; ++i) cin >> A[i];
	sort(A.begin(), A.end(), greater<i64>());
	
	auto less_count = [&](i64 x) {
		i64 cnt = 0;
		for (int i=1; i<N; ++i) {
			int lb = -1, ub = i;
			while (ub - lb > 1) {
				int med = (lb + ub) / 2;
				if ((A[i] > 0 and A[i] * A[med] > x) or (A[i] <= 0 and A[i] * A[med] <= x)) lb = med;
				else ub = med;
			}
			cnt += (A[i] > 0 ? i - ub : ub);
		}
		return cnt;
	};
		
	i64 lb = -1e18 - 1, ub = 1e18;
	while (ub - lb > 1) {
		i64 med = (lb + ub) / 2;
		if (K <= less_count(med)) ub = med;
		else lb = med;
	}
	cout << ub << endl;
}
```