#include <stdio.h>
#include <algorithm>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, mn = -1e18 - 1, mx = 1e18 + 1;
  scanf("%lld%lld", &n, &k);
  ll a[n];
  for(i = 0; i < n; ++i) scanf("%lld", &a[i]);
  sort(a, a + n);
  while(mx - mn != 1) {
    ll now = (mn + mx) / 2, cnt = 0;
    for(i = 0; i < n; ++i) {
      ll tmp = 0;
      if(a[i] < 0) {
        if(lower_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] + 1 : now / a[i]) > a + i) cnt += n - (lower_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] + 1 : now / a[i]) - a);
        else cnt += n - i - 1;
      } else if(a[i] == 0) {
        if(now >= 0) cnt += n - i - 1;
      } else {
        if(upper_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] - 1 : now / a[i]) > a + i) cnt += upper_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] - 1 : now / a[i]) - a - i - 1;
      }
    }
    if(k <= cnt) mx = now;
    else mn = now;
  }
  printf("%lld", mx);
  return 0;
}
### Prompt

Please provide a CPP coded solution 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 <stdio.h>
#include <algorithm>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, mn = -1e18 - 1, mx = 1e18 + 1;
  scanf("%lld%lld", &n, &k);
  ll a[n];
  for(i = 0; i < n; ++i) scanf("%lld", &a[i]);
  sort(a, a + n);
  while(mx - mn != 1) {
    ll now = (mn + mx) / 2, cnt = 0;
    for(i = 0; i < n; ++i) {
      ll tmp = 0;
      if(a[i] < 0) {
        if(lower_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] + 1 : now / a[i]) > a + i) cnt += n - (lower_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] + 1 : now / a[i]) - a);
        else cnt += n - i - 1;
      } else if(a[i] == 0) {
        if(now >= 0) cnt += n - i - 1;
      } else {
        if(upper_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] - 1 : now / a[i]) > a + i) cnt += upper_bound(a, a + n, (now % a[i] && now < 0) ? now / a[i] - 1 : now / a[i]) - a - i - 1;
      }
    }
    if(k <= cnt) mx = now;
    else mn = now;
  }
  printf("%lld", mx);
  return 0;
}
```