#include<bits/stdc++.h>
#define rep(i, n) for(int (i)=0 ;i <(n); ++(i))
using namespace std;
typedef long long ll;

const ll INF = (ll)1e18 + 1;

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

  sort(a.begin(), a.end());
  ll l = -INF;
  ll r = INF;
  while (r - l > 1) {
    ll mid = (l+r)/2;  

    auto ok = [&](ll x) {
      ll cnt = 0;

      rep(i, n) {
        if (a[i] < 0) {
          ll l = -1;
          ll r = n;
          while (r-l>1) {
            ll ct = (r+l)/2; 
            if (a[i] * a[ct] < x) r = ct;
            else l = ct; 
          }
          cnt += n-r;
        } else {
         ll l = -1;
         ll r = n;
         while (r-l>1) { 
           ll ct = (l+r)/2;
           if (a[i] * a[ct] < x) l = ct;
           else r = ct;
         }
         cnt += r;
        }
        if (a[i]*a[i] < x) cnt--;
      }
      cnt /= 2;
      return cnt < k;
    };

    if (ok(mid)) l = mid;
    else r = mid;
  }
  cout << l << endl;
  return 0;
}
### Prompt

Generate a cpp solution to 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>
#define rep(i, n) for(int (i)=0 ;i <(n); ++(i))
using namespace std;
typedef long long ll;

const ll INF = (ll)1e18 + 1;

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

  sort(a.begin(), a.end());
  ll l = -INF;
  ll r = INF;
  while (r - l > 1) {
    ll mid = (l+r)/2;  

    auto ok = [&](ll x) {
      ll cnt = 0;

      rep(i, n) {
        if (a[i] < 0) {
          ll l = -1;
          ll r = n;
          while (r-l>1) {
            ll ct = (r+l)/2; 
            if (a[i] * a[ct] < x) r = ct;
            else l = ct; 
          }
          cnt += n-r;
        } else {
         ll l = -1;
         ll r = n;
         while (r-l>1) { 
           ll ct = (l+r)/2;
           if (a[i] * a[ct] < x) l = ct;
           else r = ct;
         }
         cnt += r;
        }
        if (a[i]*a[i] < x) cnt--;
      }
      cnt /= 2;
      return cnt < k;
    };

    if (ok(mid)) l = mid;
    else r = mid;
  }
  cout << l << endl;
  return 0;
}
```