#include "bits/stdc++.h"
using namespace std;
const long long INF = (long long)(1e18) + 1;
int main() {
	int N;
	long long K;
	cin >> N >> K;
	vector<long long>A(N);
	for (int n = 0; n < N; ++n) {
		cin >> A[n];
	}
	//小さい順にソート
	sort(A.begin(), A.end());
	long long l = -INF;
	long long r = INF;	
	while (l + 1 < r) {
		long long x = (l + r) / 2;
		bool check = [&] {
			long long tmp = 0;
			for (int n = 0; n < N; ++n) {
				if (A[n] < 0) {
					int l = -1;
					int r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if (A[c] * A[n] < x) {
							r = c;
						}
						else {
							l = c;
						}
					}
					tmp += (long long)(N - r); 
				}
				else {
					int l = -1, r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if (A[c] * A[n] < x) l = c; else r = c;
					}
					tmp += (long long)r; 
				}
				if (A[n] * A[n] < x) {
					tmp--;
				}
			}
			tmp /= 2; 
			return tmp < K;
		}();
		if (check) {
			l = x;
		}
		else {
			r = x;
		}
	}
	cout << l << endl;
	return 0;
}
### Prompt

Please formulate 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;
const long long INF = (long long)(1e18) + 1;
int main() {
	int N;
	long long K;
	cin >> N >> K;
	vector<long long>A(N);
	for (int n = 0; n < N; ++n) {
		cin >> A[n];
	}
	//小さい順にソート
	sort(A.begin(), A.end());
	long long l = -INF;
	long long r = INF;	
	while (l + 1 < r) {
		long long x = (l + r) / 2;
		bool check = [&] {
			long long tmp = 0;
			for (int n = 0; n < N; ++n) {
				if (A[n] < 0) {
					int l = -1;
					int r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if (A[c] * A[n] < x) {
							r = c;
						}
						else {
							l = c;
						}
					}
					tmp += (long long)(N - r); 
				}
				else {
					int l = -1, r = N;
					while (l + 1 < r) {
						int c = (l + r) / 2;
						if (A[c] * A[n] < x) l = c; else r = c;
					}
					tmp += (long long)r; 
				}
				if (A[n] * A[n] < x) {
					tmp--;
				}
			}
			tmp /= 2; 
			return tmp < K;
		}();
		if (check) {
			l = x;
		}
		else {
			r = x;
		}
	}
	cout << l << endl;
	return 0;
}
```