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

long n, k;
long a[200000];

long sub(long th) {
  long sum = 0;
  for (int i = 0; i < n; ++i) {
    long v = a[i];
    if (v == 0) {
      sum += th > 0 ? n - (i + 1) : 0;
    } else if (v > 0) {
      int lo = i;
      int hi = n;
      while (hi - lo > 1) {
        int mid = (lo + hi) / 2;
        if (v * a[mid] < th) {
          lo = mid;
        } else {
          hi = mid;
        }
      }
      sum += hi - (i + 1);
    } else {
      int lo = i;
      int hi = n;
      while (hi - lo > 1) {
        int mid = (lo + hi) / 2;
        if (v * a[mid] < th) {
          hi = mid;
        } else {
          lo = mid;
        }
      }
      sum += n - hi;
    }
  }
  return sum;
}

int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  long lo = -2e18;
  long hi = 2e18;
  while (hi - lo > 1) {
    long mid = (lo + hi) / 2;
    if (sub(mid) < k) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  cout << lo << endl;
}

### Prompt

Create a solution in CPP for 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;

long n, k;
long a[200000];

long sub(long th) {
  long sum = 0;
  for (int i = 0; i < n; ++i) {
    long v = a[i];
    if (v == 0) {
      sum += th > 0 ? n - (i + 1) : 0;
    } else if (v > 0) {
      int lo = i;
      int hi = n;
      while (hi - lo > 1) {
        int mid = (lo + hi) / 2;
        if (v * a[mid] < th) {
          lo = mid;
        } else {
          hi = mid;
        }
      }
      sum += hi - (i + 1);
    } else {
      int lo = i;
      int hi = n;
      while (hi - lo > 1) {
        int mid = (lo + hi) / 2;
        if (v * a[mid] < th) {
          hi = mid;
        } else {
          lo = mid;
        }
      }
      sum += n - hi;
    }
  }
  return sum;
}

int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  long lo = -2e18;
  long hi = 2e18;
  while (hi - lo > 1) {
    long mid = (lo + hi) / 2;
    if (sub(mid) < k) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  cout << lo << endl;
}

```