#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second

const int mx=100010;
const ll mod=1e9+7;

vector<ll> a(mx);
ll n,k;

bool finish(ll x){
  ll cnt=0;
  for(int i=0; i<n; i++){
    if(a[i]<0){
      ll v = (x<0?(x-a[i]+1):x)/a[i]; 
      cnt += a.begin()+i - lower_bound(a.begin(),a.begin()+i, v);
    } else if(a[i]==0){
      if(x>=0){ cnt += i; }
    } else {
      ll v = (x>=0?x:(x-a[i]+1))/a[i];
      cnt += upper_bound(a.begin(),a.begin()+i, v) - a.begin();
    }
  }
  return cnt>=k;
}
 
int main(){
  cin >> n >> k;
  a.resize(n);
  rep(i,n){cin>>a[i];}
  sort(a.begin(),a.end());
  ll l=-longinf,r=longinf,mid=(r+l)/2;
  while(r-l>1){
    mid=(r+l)/2;
    if(finish(mid)){ r=mid; }else{ l=mid; }
  }
  ll ans = r;
  cout << ans << endl;
  return 0;
}

### 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 pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second

const int mx=100010;
const ll mod=1e9+7;

vector<ll> a(mx);
ll n,k;

bool finish(ll x){
  ll cnt=0;
  for(int i=0; i<n; i++){
    if(a[i]<0){
      ll v = (x<0?(x-a[i]+1):x)/a[i]; 
      cnt += a.begin()+i - lower_bound(a.begin(),a.begin()+i, v);
    } else if(a[i]==0){
      if(x>=0){ cnt += i; }
    } else {
      ll v = (x>=0?x:(x-a[i]+1))/a[i];
      cnt += upper_bound(a.begin(),a.begin()+i, v) - a.begin();
    }
  }
  return cnt>=k;
}
 
int main(){
  cin >> n >> k;
  a.resize(n);
  rep(i,n){cin>>a[i];}
  sort(a.begin(),a.end());
  ll l=-longinf,r=longinf,mid=(r+l)/2;
  while(r-l>1){
    mid=(r+l)/2;
    if(finish(mid)){ r=mid; }else{ l=mid; }
  }
  ll ans = r;
  cout << ans << endl;
  return 0;
}

```