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

int main() {
  int64_t n, k;
  cin >> n >> k;
  vector<int64_t> a(n);
  for (int i = 0; i < n; i++) cin >> a.at(i);
  sort(a.begin(), a.end());
  int64_t lowest = -2e18, highest = 2e18;
  for (;;) {
    int64_t half = (highest + lowest) / 2, count = 0; // half以下になるペアの個数
    for (int i = 0; i < n - 1; i++) {
      if (a.at(i) < 0) {
        int64_t r = half / a.at(i);
        if (r * a.at(i) > half) r++;
        count += a.end() - lower_bound(a.begin() + i + 1, a.end(), r);
      } else if (a.at(i) == 0) {
        if (half >= 0) count += n - i - 1;
      } else {
        int64_t r = half / a.at(i);
        if (r * a.at(i) > half) r--;
        count += upper_bound(a.begin() + i + 1, a.end(), r) - (a.begin() + i + 1);
      }
    }
    if (count < k) {
      lowest = half;
    } else {
      highest = half;
    }
    if (highest - lowest == 1) break;
  }
  cout << highest << endl;
}

### 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() {
  int64_t n, k;
  cin >> n >> k;
  vector<int64_t> a(n);
  for (int i = 0; i < n; i++) cin >> a.at(i);
  sort(a.begin(), a.end());
  int64_t lowest = -2e18, highest = 2e18;
  for (;;) {
    int64_t half = (highest + lowest) / 2, count = 0; // half以下になるペアの個数
    for (int i = 0; i < n - 1; i++) {
      if (a.at(i) < 0) {
        int64_t r = half / a.at(i);
        if (r * a.at(i) > half) r++;
        count += a.end() - lower_bound(a.begin() + i + 1, a.end(), r);
      } else if (a.at(i) == 0) {
        if (half >= 0) count += n - i - 1;
      } else {
        int64_t r = half / a.at(i);
        if (r * a.at(i) > half) r--;
        count += upper_bound(a.begin() + i + 1, a.end(), r) - (a.begin() + i + 1);
      }
    }
    if (count < k) {
      lowest = half;
    } else {
      highest = half;
    }
    if (highest - lowest == 1) break;
  }
  cout << highest << endl;
}

```