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

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  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 ok = -2e18, ng = 2e18;
  while (ng - ok > 1) {
    long mid = (ok + ng) / 2, sum = 0;
    for (int i = 0; i < N; i++) {
      long double tmp = (long double) mid / A.at(i);
      if (A.at(i) > 0) sum += lower_bound(A.begin(), A.end(), tmp) - A.begin();
      if (A.at(i) < 0) sum += N - (upper_bound(A.begin(), A.end(), tmp) - A.begin());
      if (A.at(i) == 0) sum += (mid > 0) ? N : 0;
      if (A.at(i) * A.at(i) < mid) sum--;
    }
    (sum / 2 < K) ? ok = mid : ng = mid;
  }
  cout << ok << "\n";
}

### 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;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  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 ok = -2e18, ng = 2e18;
  while (ng - ok > 1) {
    long mid = (ok + ng) / 2, sum = 0;
    for (int i = 0; i < N; i++) {
      long double tmp = (long double) mid / A.at(i);
      if (A.at(i) > 0) sum += lower_bound(A.begin(), A.end(), tmp) - A.begin();
      if (A.at(i) < 0) sum += N - (upper_bound(A.begin(), A.end(), tmp) - A.begin());
      if (A.at(i) == 0) sum += (mid > 0) ? N : 0;
      if (A.at(i) * A.at(i) < mid) sum--;
    }
    (sum / 2 < K) ? ok = mid : ng = mid;
  }
  cout << ok << "\n";
}

```