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 <algorithm>
#include <iostream>
#include <vector>

using ll = long long int;
using namespace std;

int main(){
    int n, k; cin >> n >> k;
    vector<ll> pluss;
    vector<ll> minuss;
    vector<ll> zeros;
    for(int i=0; i<n; ++i){
        ll a; cin >> a;
        if(a < 0){ minuss.push_back(a); }
        else if(a == 0){ zeros.push_back(a); }
        else{ pluss.push_back(a); }
    }
    sort(pluss.begin(), pluss.end());
    sort(minuss.begin(), minuss.end());

    ll p_len = pluss.size();
    ll m_len = minuss.size();
    ll z_len = zeros.size();

    ll minus_num = p_len * m_len;
    ll zeros_num = z_len * (z_len - 1) / 2 + z_len * (m_len + p_len);

    ll l = -(ll)1e18 - 10LL; ll r = (ll)1e18 + 10LL;
    while(r - l > 1LL){
        ll mid = (l + r) / 2LL;
        ll cnt = 0;
        if(mid > 0LL){
            ll ret = 0;
            for(int i=0; i<(int)pluss.size(); ++i){
                ret += upper_bound(pluss.begin(), pluss.end(), mid / pluss[i]) - pluss.begin(); 
                if(pluss[i] * pluss[i] <= mid){ ret -= 1LL; }
            }
            cnt += ret / 2LL;
            ret = 0;
            for(int i=0; i<(int)minuss.size(); ++i){
                ret += m_len - (int)(lower_bound(minuss.begin(), minuss.end(), mid / minuss[i]) - minuss.begin()); 
                if(minuss[i] * minuss[i] <= mid){ ret -= 1LL; }
            }
            cnt += ret / 2LL;
            cnt += minus_num;
            cnt += zeros_num;
        }
        else if(mid == 0LL){
            cnt += minus_num;
            cnt += zeros_num;
        }
        else{
            for(int i=0; i<(int)pluss.size(); ++i){
                cnt += upper_bound(minuss.begin(), minuss.end(), (mid - pluss[i] + 1) / pluss[i]) - minuss.begin();
            }
        }
        if (cnt >= k){
            r = mid;
        }
        else{
            l = mid;
        }
    }
    cout << r << endl;
}