# include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define chmin(a, b) a = min(a, b)
#define chmax(a, b) a = max(a, b)
using ll = long long;
const ll INF = ll(1e18) + 1;

bool search(int n, ll k, vector<ll> a, ll mid)
{
  ll tot = 0;
  rep(i, n) {
    int l = -1, r = n;

    if(a[i] < 0) {
      while(l+1 < r) {
        int m = (l+r) / 2;
        if (a[m] * a[i] < mid) r = m; else l = m;
      }
      tot += n - r;
    } else {
      while(l+1 < r) {
        int m = (l+r) / 2;
        if (a[m] * a[i] < mid) l = m; else r = m;
      }
      tot += r;
    }
    if (a[i] * a[i] < mid) tot--;
  }
  return (tot / 2 < k);
}

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 mid = (l + r) / 2;
    
    // K未満の値が見つかるかをチェック
    if (search(n, k, a, mid)) l = mid; else r = mid; 
  }
  
  cout << l << endl;
  return 0;
}
### Prompt

Develop a solution in Cpp 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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define chmin(a, b) a = min(a, b)
#define chmax(a, b) a = max(a, b)
using ll = long long;
const ll INF = ll(1e18) + 1;

bool search(int n, ll k, vector<ll> a, ll mid)
{
  ll tot = 0;
  rep(i, n) {
    int l = -1, r = n;

    if(a[i] < 0) {
      while(l+1 < r) {
        int m = (l+r) / 2;
        if (a[m] * a[i] < mid) r = m; else l = m;
      }
      tot += n - r;
    } else {
      while(l+1 < r) {
        int m = (l+r) / 2;
        if (a[m] * a[i] < mid) l = m; else r = m;
      }
      tot += r;
    }
    if (a[i] * a[i] < mid) tot--;
  }
  return (tot / 2 < k);
}

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 mid = (l + r) / 2;
    
    // K未満の値が見つかるかをチェック
    if (search(n, k, a, mid)) l = mid; else r = mid; 
  }
  
  cout << l << endl;
  return 0;
}
```