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 <bits/stdc++.h>

using namespace std;
using i64 = int64_t;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
const int INF = (1 << 30);
const i64 INFL = (1LL << 62);
const i64 MOD = 1000000007;

bool check(i64 x, i64 k, const vector<i64> &a) {
  // x 以下の数がk個以上ある？
  i64 cnt = 0;  // x以下の数の個数
  i64 n = a.size();

  rep(i, n) {
    i64 ci = 0;
    double t = 0;
    if (a[i] == 0) {
      // 0とかけてx以下になる個数
      if (x < 0) {
      } else {
        ci = n - 1 - i;
      }
    } else {
      t = x / a[i];
      if (a[i] > 0) {
        // t以下の数の個数
        ci = upper_bound(a.begin() + i + 1, a.end(), t) - (a.begin() + i + 1);
      } else {
        // t以上の数の個数
        ci = a.end() - lower_bound(a.begin() + i + 1, a.end(), t);
      }
    }
    cnt += ci;
    // cout << "a[i]:" << a[i] << ", x:" << x << ", t:" << t << ", ci:" << ci
    //      << endl;
  }

  return cnt >= k;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  i64 n, k;
  cin >> n >> k;
  vector<i64> a(n);
  rep(i, n) { cin >> a[i]; }
  sort(all(a));
  const i64 MAX = 1e18 * 2;
  i64 ok = MAX;
  i64 ng = -MAX;
  while (abs(ok - ng) > 1) {
    i64 mid = (ok + ng) / 2;
    if (check(mid, k, a)) {
      ok = mid;
    } else {
      ng = mid;
    }
  }
  cout << ok << endl;
  return 0;
}