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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

int n;
long long k;
vector<long long> v;
vector<long long> vpos;

bool check(long long x) {
    long long cnt = 0;
    if (v[0] < 0) {
        int i = 0;
        int j;
        for (j = 0; j < n; j++) {
            if (v[j] * v[i] <= x) {
                break;
            }
        }
        while (i >= j) {
            j++;
        }
        cnt += n-j;
        for (i = 1; i < n; i++) {
            if (v[i] >= 0) {
                break;
            }
            while (j > i && v[i] * v[j] <= x) {
                j--;
            }
            j++;
            while (i >= j) {
                j++;
            }
            cnt += n-j;
            //cnt += n-j;
            //if (i >= j) {
            //    cnt--;
            //}
        }
    }
    //cout << "negcnt " << cnt  << endl;
    int np = vpos.size();
    if (np > 0) {
        int i = np-1;
        int j;
        for (j = np-1; j >= 0; j--) {
            if (vpos[j] * vpos[i] <= x) {
                break;
            }
        }
        while (i <= j) {
            j--;
        }
        cnt += j+1;
        //cout << v[i] << " add " << j+1 << endl;
        for (i = np-2; i >= 0; i--) {
            if (vpos[i] < 0) {
                break;
            }
            while (j < i && vpos[i] * vpos[j] <= x) {
                j++;
            }
            j--;
            while (i <= j) {
                j--;
            }
            cnt += j+1;
            //cout << v[i] << " add " << j+1 << endl;
            //cnt += j+1;
            //if (i <= j) {
            //    cnt--;
            //}
        }
    }
    //cout << "HAD " << cnt << " of " << k << endl;
    return cnt >= k;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    cin >> n;
    cin >> k;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        v.push_back(x);
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < n; i++) {
        if (v[i] >= 0) {
            vpos.push_back(v[i]);
        }
    }

    long long INF = 2e18;
    long long l = -INF;
    long long r = INF;
    while (l+1 != r) {
        long long mid = (l+r)/2;
        //cout << l << " " << r << " " << mid << endl;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid;
        }
    }
    cout << r << endl;

    return 0;
}