#include <bits/stdc++.h>
using namespace std;
typedef int64_t ll;
const ll INF=(ll)1e18+1;

int main() {
  ll n,k,p=0,z=0,m=0,mid,s,ri;
  cin >> n >> k;
  ll a[n];
  for(int i=0;i<n;i++){
    cin >> a[i];
    (a[i]<0 ? m : (a[i]>0 ? p : z))++;
  }
  sort(a,a+n);
  ll ok=INF,ng=-INF;
  while(ok-ng>1){
    mid=ng+(ok-ng)/2; //mid未満の積はいくつあるか
    if(mid==0) s=m*p;
    if(mid>0){
      s=m*p + z*(m+p) + z*(z-1)/2;
      ri=n-1;
      for(int i=m+z;i<n;i++){
        while(a[i]*a[ri]>=mid) ri--;
        if(ri>i) s+=ri-i;
        else break;
      }
      ri=0;
      for(int i=m-1;i>=0;i--){
        while(a[i]*a[ri]>=mid) ri++;
        if(ri<i) s+=i-ri;
        else break;
      }
    }
    if(mid<0){
      s=0;
      ri=m+z;
      for(int i=0;i<m;i++){
        while(a[i]*a[ri]>=mid) ri++;
        if(ri<n) s+=n-ri;
        else break;
      }
    }
    if(s<k)ng=mid;
    else ok=mid;
  }
  cout << ng << endl;
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
typedef int64_t ll;
const ll INF=(ll)1e18+1;

int main() {
  ll n,k,p=0,z=0,m=0,mid,s,ri;
  cin >> n >> k;
  ll a[n];
  for(int i=0;i<n;i++){
    cin >> a[i];
    (a[i]<0 ? m : (a[i]>0 ? p : z))++;
  }
  sort(a,a+n);
  ll ok=INF,ng=-INF;
  while(ok-ng>1){
    mid=ng+(ok-ng)/2; //mid未満の積はいくつあるか
    if(mid==0) s=m*p;
    if(mid>0){
      s=m*p + z*(m+p) + z*(z-1)/2;
      ri=n-1;
      for(int i=m+z;i<n;i++){
        while(a[i]*a[ri]>=mid) ri--;
        if(ri>i) s+=ri-i;
        else break;
      }
      ri=0;
      for(int i=m-1;i>=0;i--){
        while(a[i]*a[ri]>=mid) ri++;
        if(ri<i) s+=i-ri;
        else break;
      }
    }
    if(mid<0){
      s=0;
      ri=m+z;
      for(int i=0;i<m;i++){
        while(a[i]*a[ri]>=mid) ri++;
        if(ri<n) s+=n-ri;
        else break;
      }
    }
    if(s<k)ng=mid;
    else ok=mid;
  }
  cout << ng << endl;
}

```