Problem:
Score: 400 points

Problem Statement

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.

Sample Input 1

4 3
3 3 -4 -2

Sample Output 1

-6

There are six ways to form a pair. The products of those pairs are 9, -12, -6, -12, -6, 8.

Sorting those numbers in ascending order, we have -12, -12, -6, -6, 8, 9. The third number in this list is -6.

Sample Input 2

10 40
5 4 3 2 -1 0 0 0 0 0

Sample Output 2

6

Sample Input 3

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

Sample Output 3

448283280358331064

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <queue>
#include <map>
#include <numeric>
#include <unordered_map>
#include <iomanip>
#include <functional>
#include <bitset>
#include <complex>
#include <stack>
#include <cstdint>



#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define rrep(i, n) for(ll i = (ll)(n-1); i >= 0; i--)
#define repi(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rrepi(i,a,b) for(ll i=(ll)(b-1);i>=(ll)(a);i--)

#define all(x) (x).begin(),(x).end()

template<class T>inline bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>inline bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

typedef long long ll;

using namespace std;

ll n,k;
vector<ll>a;
ll lo(ll num,ll d){
  ll mo=num%d;
  if(mo<0)mo+=d;
  return (num-mo)/d;
}
ll hi(ll num,ll d){

  ll mo=num%d;
  if(mo<0)mo+=d;
  if(mo!=0)mo-=d;
  return (num-mo)/d;
}
bool check(ll num){//numがkばんめ以降か
  ll cnt=0;//numより小さいもの
  rep(i,n){
    if(a[i]==0){
      if(0ll<num)cnt+=n-1-i;
      continue;
    }
    if(a[i]>0){
      ll tn=hi(num,a[i]);
      ll id=lower_bound(all(a),tn)-a.begin();
      cnt+=max(0ll,id-i-1);
    }
    else {
      ll tn=lo(num,a[i]);
      ll id=upper_bound(all(a),tn)-a.begin();
      cnt+=max(0ll,n-max(i+1,id));
    }
  }
  //cout<<cnt<<" "<<num<<endl;
  return cnt>=k;
}
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin>>n>>k;
  a.resize(n);
  rep(i,n)cin>>a[i];
  sort(all(a));
  ll ng=-(ll)1e18-7,ok=(ll)1e18+7;

  while(abs(ok-ng)>1){
    ll mid=(ok+ng)/2;
    if(check(mid))ok=mid;
    else ng=mid;
  }
  cout<<ng<<endl;

  return 0;

}