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

int main() {
  long N, K;
  cin >> N >> K;
  vector<long> A(N);
  for (int i = 0; i < N; i++) cin >> A.at(i);
  sort(A.begin(), A.end());

  long L = -2e18, R = 2e18;
  while (R - L > 1) {
    long M = (L + R) / 2, cnt = 0;
    for (int i = 0; i < N; i++) {
      if (A.at(i) < 0) {
        int l = i, r = N;
        while (r - l > 1) {
          int m = (l + r) / 2;
          (A.at(i) * A.at(m) < M) ? r = m : l = m;
        }
        cnt += N - r;
      } else {
        int l = i, r = N;
        while (r - l > 1) {
          int m = (l + r) / 2;
          (A.at(i) * A.at(m) < M) ? l = m : r = m;
        }
        cnt += l - i;
      }
    }
    (cnt < K) ? L = M : R = M;
  }
  cout << L << "\n";
}
### Prompt

Develop a solution in cpp 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 N, K;
  cin >> N >> K;
  vector<long> A(N);
  for (int i = 0; i < N; i++) cin >> A.at(i);
  sort(A.begin(), A.end());

  long L = -2e18, R = 2e18;
  while (R - L > 1) {
    long M = (L + R) / 2, cnt = 0;
    for (int i = 0; i < N; i++) {
      if (A.at(i) < 0) {
        int l = i, r = N;
        while (r - l > 1) {
          int m = (l + r) / 2;
          (A.at(i) * A.at(m) < M) ? r = m : l = m;
        }
        cnt += N - r;
      } else {
        int l = i, r = N;
        while (r - l > 1) {
          int m = (l + r) / 2;
          (A.at(i) * A.at(m) < M) ? l = m : r = m;
        }
        cnt += l - i;
      }
    }
    (cnt < K) ? L = M : R = M;
  }
  cout << L << "\n";
}
```