#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
#define rep(i,N) for(ll (i)=0;(i)<(N);(i)++)
const int mod = 1000000007;
const ll INF = ll(1e18) + 1;

int main() {
  int n;
  ll k;
  cin >> n >> k;
  vector<ll> a(n);
  rep(i, n) cin >> a[i];
  sort(a.begin(), a.end());

  ll l = -INF, r = INF;
  while (l + 1 < r) {
    ll x = (l + r) / 2;

    bool ok = [&] {
      ll total = 0;
      rep(i, n) {
        int l = -1, r = n;
        if(a[i] >= 0) {
          while (l + 1 < r) {
            int c = (l + r) / 2;
            if(a[i] * a[c] < x) l = c; else r = c;
          }
          total += l + 1;
        } else {
          while (l + 1 < r) {
            int c = (l + r) / 2;
            if(a[i] * a[c] < x) r = c; else l = c;
          }
          total += n - r;
        }
        if(a[i] * a[i] < x) total--;
      }
      total /= 2;
      return total < k;
    }();
    if(ok) l = x;
    else r = x;
  }
  cout << l << 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;
typedef long long ll;
typedef pair<ll,ll> P;
#define rep(i,N) for(ll (i)=0;(i)<(N);(i)++)
const int mod = 1000000007;
const ll INF = ll(1e18) + 1;

int main() {
  int n;
  ll k;
  cin >> n >> k;
  vector<ll> a(n);
  rep(i, n) cin >> a[i];
  sort(a.begin(), a.end());

  ll l = -INF, r = INF;
  while (l + 1 < r) {
    ll x = (l + r) / 2;

    bool ok = [&] {
      ll total = 0;
      rep(i, n) {
        int l = -1, r = n;
        if(a[i] >= 0) {
          while (l + 1 < r) {
            int c = (l + r) / 2;
            if(a[i] * a[c] < x) l = c; else r = c;
          }
          total += l + 1;
        } else {
          while (l + 1 < r) {
            int c = (l + r) / 2;
            if(a[i] * a[c] < x) r = c; else l = c;
          }
          total += n - r;
        }
        if(a[i] * a[i] < x) total--;
      }
      total /= 2;
      return total < k;
    }();
    if(ok) l = x;
    else r = x;
  }
  cout << l << endl;

}
```