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

ll n, k;
vector<ll> a;
ll INF = (ll)(1e18)+1;

ll judge(ll x) {
  ll cnt = 0;
  rep(i, n) {
    if (a.at(i) < 0) {
      int l = -1;
      int r = n;
      while (r-l>1) {
        int c = (r+l)/2;
        if (a.at(i)*a.at(c)<x) r = c;
        else l = c;
      }
      cnt += n-r;
    }
    else {
      ll l = -1;
      ll r = n;
      while (r-l>1) {
        int c = (r+l)/2;
        if (a.at(i)*a.at(c)<x) l = c;
        else r = c;
      }
      cnt += r;
    }
    if (a.at(i)*a.at(i)<x) cnt--;
  } 
  cnt /= 2;
  return cnt;
}

int main() {
  cin >> n >> k;
  a.resize(n);
  rep(i, n) cin >> a.at(i);
  sort(a.begin(), a.end());

  ll l = -INF;
  ll r = INF;
  while (r-l>1) {
    ll c = (r+l)/2; 
    if (judge(c)<k) l = c;
    else r = c;
  }
  cout << l << 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<bits/stdc++.h>
#define rep(i, n) for(int i=0 ;i <(n); ++i)
using namespace std;
typedef long long ll;

ll n, k;
vector<ll> a;
ll INF = (ll)(1e18)+1;

ll judge(ll x) {
  ll cnt = 0;
  rep(i, n) {
    if (a.at(i) < 0) {
      int l = -1;
      int r = n;
      while (r-l>1) {
        int c = (r+l)/2;
        if (a.at(i)*a.at(c)<x) r = c;
        else l = c;
      }
      cnt += n-r;
    }
    else {
      ll l = -1;
      ll r = n;
      while (r-l>1) {
        int c = (r+l)/2;
        if (a.at(i)*a.at(c)<x) l = c;
        else r = c;
      }
      cnt += r;
    }
    if (a.at(i)*a.at(i)<x) cnt--;
  } 
  cnt /= 2;
  return cnt;
}

int main() {
  cin >> n >> k;
  a.resize(n);
  rep(i, n) cin >> a.at(i);
  sort(a.begin(), a.end());

  ll l = -INF;
  ll r = INF;
  while (r-l>1) {
    ll c = (r+l)/2; 
    if (judge(c)<k) l = c;
    else r = c;
  }
  cout << l << endl;
  return 0;
}
```