#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using namespace std;
using i64 = int64_t;

bool check(i64 x, i64 k, const vector<i64> &a) {
  i64 cnt = 0;
  i64 n = a.size();
 
  rep(i, n) {
    i64 ci = 0;
    long double t = 0;
    if (a[i] == 0) {
      if (x >= 0) {
        ci = n - 1 - i;
      }
    } else {
      t = x / (long double)a[i];
      if (a[i] > 0) {
        ci = upper_bound(a.begin() + i + 1, a.end(), t) - (a.begin() + i + 1);
      } else {
        ci = a.end() - lower_bound(a.begin() + i + 1, a.end(), t);
      }
    }
    cnt += ci;
  }
    return cnt >= k;
}

int main(){
    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;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define rep(i, n) for(int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using namespace std;
using i64 = int64_t;

bool check(i64 x, i64 k, const vector<i64> &a) {
  i64 cnt = 0;
  i64 n = a.size();
 
  rep(i, n) {
    i64 ci = 0;
    long double t = 0;
    if (a[i] == 0) {
      if (x >= 0) {
        ci = n - 1 - i;
      }
    } else {
      t = x / (long double)a[i];
      if (a[i] > 0) {
        ci = upper_bound(a.begin() + i + 1, a.end(), t) - (a.begin() + i + 1);
      } else {
        ci = a.end() - lower_bound(a.begin() + i + 1, a.end(), t);
      }
    }
    cnt += ci;
  }
    return cnt >= k;
}

int main(){
    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;
}

```