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

int main() {
	long long N, K;
	vector<long long> A;
	cin >> N >> K;
	A.resize(N);
	for (int i = 0; i < N; i++) cin >> A[i];
	sort(A.begin(), A.end());
	long long L = -1000000000000000001, R = 1000000000000000000;
	while (R - L > 1) {
		long long MID = (L + R) / 2;
		long long COUNT = 0;
		for (int i = 0; i < N; i++) {
			if (A[i] < 0) {
				long long L2 = -1, R2 = N;
				while (R2 - L2 > 1) {
					long long MID2 = (L2 + R2) / 2;
					if (A[i] * A[MID2] > MID) L2 = MID2;
					else R2 = MID2;
				}
				COUNT += N - R2;
			}
			if (A[i] > 0) {
				long long L2 = -1, R2 = N;
				while (R2 - L2 > 1) {
					long long MID2 = (L2 + R2) / 2;
					if (A[i] * A[MID2] <= MID) L2 = MID2;
					else R2 = MID2;
				}
				COUNT += R2;
			}
			if (A[i] == 0) {
				if (MID >= 0) COUNT += N;
			}
			if (A[i] * A[i] <= MID) COUNT--;
		}
		COUNT /= 2;
		if (COUNT < K) L = MID;
		else R = MID;
	}
	cout << R << endl;
}
### 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;

int main() {
	long long N, K;
	vector<long long> A;
	cin >> N >> K;
	A.resize(N);
	for (int i = 0; i < N; i++) cin >> A[i];
	sort(A.begin(), A.end());
	long long L = -1000000000000000001, R = 1000000000000000000;
	while (R - L > 1) {
		long long MID = (L + R) / 2;
		long long COUNT = 0;
		for (int i = 0; i < N; i++) {
			if (A[i] < 0) {
				long long L2 = -1, R2 = N;
				while (R2 - L2 > 1) {
					long long MID2 = (L2 + R2) / 2;
					if (A[i] * A[MID2] > MID) L2 = MID2;
					else R2 = MID2;
				}
				COUNT += N - R2;
			}
			if (A[i] > 0) {
				long long L2 = -1, R2 = N;
				while (R2 - L2 > 1) {
					long long MID2 = (L2 + R2) / 2;
					if (A[i] * A[MID2] <= MID) L2 = MID2;
					else R2 = MID2;
				}
				COUNT += R2;
			}
			if (A[i] == 0) {
				if (MID >= 0) COUNT += N;
			}
			if (A[i] * A[i] <= MID) COUNT--;
		}
		COUNT /= 2;
		if (COUNT < K) L = MID;
		else R = MID;
	}
	cout << R << endl;
}
```