#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const ll INF = ll(1e18)+1;
ll n, k;

vector<ll> a;
bool f(ll x){
   ll count = 0;
   for(int i = 0; i < n; i++){
       if(a[i] < 0){
           int l = -1, r = n;
           while(l + 1 < r){
               int c = (l+r)/2;
               if(a[c] * a[i] < x) r = c;
               else l = c;
           }
           count += n - r;
       }
       else{
           int l = -1, r = n;
           while(l + 1 < r){
               int c = (l+r)/2;
               if(a[c] * a[i] < x) l = c;
               else r = c;
           }
           count += r;
       }
       if(a[i] * a[i] < x) count--;
   }
   count /= 2;
   return count < k;
}

int main(){
    cin >> n >> k;
    a.resize(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    sort(a.begin(), a.end());

    ll l = -INF, r = INF;
    while(l + 1 < r){
        ll x = (l+r)/2;
        if(f(x)) l = x;
        else r = x;
    }
    cout << l << endl;
}
### Prompt

Please formulate 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>
typedef long long ll;
using namespace std;

const ll INF = ll(1e18)+1;
ll n, k;

vector<ll> a;
bool f(ll x){
   ll count = 0;
   for(int i = 0; i < n; i++){
       if(a[i] < 0){
           int l = -1, r = n;
           while(l + 1 < r){
               int c = (l+r)/2;
               if(a[c] * a[i] < x) r = c;
               else l = c;
           }
           count += n - r;
       }
       else{
           int l = -1, r = n;
           while(l + 1 < r){
               int c = (l+r)/2;
               if(a[c] * a[i] < x) l = c;
               else r = c;
           }
           count += r;
       }
       if(a[i] * a[i] < x) count--;
   }
   count /= 2;
   return count < k;
}

int main(){
    cin >> n >> k;
    a.resize(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    sort(a.begin(), a.end());

    ll l = -INF, r = INF;
    while(l + 1 < r){
        ll x = (l+r)/2;
        if(f(x)) l = x;
        else r = x;
    }
    cout << l << endl;
}
```