#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 2e18;

int main(){
  ll n, k;
  cin >> n >> k;
  ll a[n];
  rep(i,n) cin >> a[i];
  sort(a,a+n);
  ll l = -INF, r = INF;
  while(r - l > 1) {
    ll m = (l + r)/2;
    ll t = 0;
    rep(i,n) {
      if(a[i] == 0) {
        if(m >= 0) t += n; 
      } else if(a[i] > 0) {
        int b;
        if(m >= 0) {
          b = upper_bound(a,a+n,m/a[i]) - a;
        } else {
          ll x = -m;
          ll c = x/a[i];
          if(x % a[i]) c++;
          b = upper_bound(a,a+n,-c) - a;
        }
        t += b;
      } else {
        int b;
        if(m < 0) {
          ll x = -m, y = -a[i];
          b = lower_bound(a,a+n,(x+y-1)/y) - a; 
        } else {
          ll y = -a[i];
          b = lower_bound(a,a+n,-(m/y)) - a; 
        }
        t += n - b;
      }
      if(a[i]*a[i] <= m) t--;
    }
    if(t >= 2*k) r = m;
    else l = m;
  } 
  cout << r << 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>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 2e18;

int main(){
  ll n, k;
  cin >> n >> k;
  ll a[n];
  rep(i,n) cin >> a[i];
  sort(a,a+n);
  ll l = -INF, r = INF;
  while(r - l > 1) {
    ll m = (l + r)/2;
    ll t = 0;
    rep(i,n) {
      if(a[i] == 0) {
        if(m >= 0) t += n; 
      } else if(a[i] > 0) {
        int b;
        if(m >= 0) {
          b = upper_bound(a,a+n,m/a[i]) - a;
        } else {
          ll x = -m;
          ll c = x/a[i];
          if(x % a[i]) c++;
          b = upper_bound(a,a+n,-c) - a;
        }
        t += b;
      } else {
        int b;
        if(m < 0) {
          ll x = -m, y = -a[i];
          b = lower_bound(a,a+n,(x+y-1)/y) - a; 
        } else {
          ll y = -a[i];
          b = lower_bound(a,a+n,-(m/y)) - a; 
        }
        t += n - b;
      }
      if(a[i]*a[i] <= m) t--;
    }
    if(t >= 2*k) r = m;
    else l = m;
  } 
  cout << r << endl;
  return 0;
}
```