#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(x) (x).begin(), (x).end()
#define dbg(x) cerr << #x << ": " << x << endl

const ll INF = (ll)(1e18) + 1;
const int MAXN = 200005;

vector<ll> a(MAXN);

bool f(ll x, ll k) {
  ll sum = 0LL;
  int n = a.size();
  rep(i, n) {
    int ng = a[i] < 0 ? -1 : n;
    int ok = a[i] < 0 ? n : -1;
    while(abs(ok - ng) > 1) {
      int mid = (ng + ok) / 2;
      if(a[mid] * a[i] < x)
        ok = mid;
      else
        ng = mid;
    }
    sum += a[i] < 0 ? n - ok : ok + 1;
  }
  rep(i, n) if(a[i] * a[i] < x) sum--;
  sum >>= 1;
  return sum < k;
}

int main() {
  int n;
  ll k;
  cin >> n >> k;
  a.resize(n);
  rep(i, n) cin >> a[i];
  sort(all(a));
  ll ok = -INF, ng = INF;
  while(abs(ok - ng) > 1) {
    ll mid = (ok + ng) / 2;
    if(f(mid, k))
      ok = mid;
    else
      ng = mid;
  }
  cout << ok << endl;
  return 0;
}
### 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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(x) (x).begin(), (x).end()
#define dbg(x) cerr << #x << ": " << x << endl

const ll INF = (ll)(1e18) + 1;
const int MAXN = 200005;

vector<ll> a(MAXN);

bool f(ll x, ll k) {
  ll sum = 0LL;
  int n = a.size();
  rep(i, n) {
    int ng = a[i] < 0 ? -1 : n;
    int ok = a[i] < 0 ? n : -1;
    while(abs(ok - ng) > 1) {
      int mid = (ng + ok) / 2;
      if(a[mid] * a[i] < x)
        ok = mid;
      else
        ng = mid;
    }
    sum += a[i] < 0 ? n - ok : ok + 1;
  }
  rep(i, n) if(a[i] * a[i] < x) sum--;
  sum >>= 1;
  return sum < k;
}

int main() {
  int n;
  ll k;
  cin >> n >> k;
  a.resize(n);
  rep(i, n) cin >> a[i];
  sort(all(a));
  ll ok = -INF, ng = INF;
  while(abs(ok - ng) > 1) {
    ll mid = (ok + ng) / 2;
    if(f(mid, k))
      ok = mid;
    else
      ng = mid;
  }
  cout << ok << endl;
  return 0;
}
```