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

const long inf = 2e18;
long n, k;
long a[200000];

template <class T>
long bin(long lo, long hi, T f, bool rev) {
  while (hi - lo > 1) {
    long mid = (lo + hi) / 2;
    if (f(mid) == rev) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  return hi;
}

bool sub(long x) {
  long sum = 0;
  for (int i = 0; i < n; ++i) {
    long v = a[i];
    auto cat = [v, x](long j) {
      return v * a[j] <= x;
    };
    
    if (v >= 0) {
      sum += bin(i, n, cat, true) - (i + 1);
    } else {
      sum += n - bin(i, n, cat, false);
    }
  }
  return sum >= k;
}

int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a, a + n);
  cout << bin(-inf, inf, sub, false) << endl;
}
### Prompt

In cpp, your task is to solve 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 inf = 2e18;
long n, k;
long a[200000];

template <class T>
long bin(long lo, long hi, T f, bool rev) {
  while (hi - lo > 1) {
    long mid = (lo + hi) / 2;
    if (f(mid) == rev) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  return hi;
}

bool sub(long x) {
  long sum = 0;
  for (int i = 0; i < n; ++i) {
    long v = a[i];
    auto cat = [v, x](long j) {
      return v * a[j] <= x;
    };
    
    if (v >= 0) {
      sum += bin(i, n, cat, true) - (i + 1);
    } else {
      sum += n - bin(i, n, cat, false);
    }
  }
  return sum >= k;
}

int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a, a + n);
  cout << bin(-inf, inf, sub, false) << endl;
}
```