#include <iostream>
#include <algorithm>

using namespace std;

#define int long long
#define rep(i,n) for(int i=0; i<(int)(n); i++)

const int N = (int)2e5;
int n;
int a[N];

int di(int x, int a) {
  return x / a + (x % a < 0 ? -1 : 0); 
}

int count(int x, int i) {
  int s = 0;
  if(a[i] > 0) s = upper_bound(a+i+1, a+n, di(x, a[i]))-a-i-1;
  else if(a[i] < 0) s = n-i-1 - (lower_bound(a+i+1, a+n, -di(x, -a[i]))-a-i-1);
  else if(x >= 0) s = n-i-1;
  return s;
}

int index(int x) {
  int s = 0;
  rep(i,n-1) s += count(x, i);
  return s;
}

int solve(int k) {
  int l = -(int)1e18-1, r = (int)1e18+1;
  while(r - l > 1) {
    int m = (l + r) / 2;
    if(index(m) < k) l = m;
    else r = m;
  }
  return r;
}

signed main() {
  int k;
  cin >> n >> k;
  rep(i,n) cin >> a[i];
  sort(a, a+n);
  
  cout << solve(k) << endl;
  return 0;
}
### Prompt

In cpp, your task is to solve the following problem:

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 <iostream>
#include <algorithm>

using namespace std;

#define int long long
#define rep(i,n) for(int i=0; i<(int)(n); i++)

const int N = (int)2e5;
int n;
int a[N];

int di(int x, int a) {
  return x / a + (x % a < 0 ? -1 : 0); 
}

int count(int x, int i) {
  int s = 0;
  if(a[i] > 0) s = upper_bound(a+i+1, a+n, di(x, a[i]))-a-i-1;
  else if(a[i] < 0) s = n-i-1 - (lower_bound(a+i+1, a+n, -di(x, -a[i]))-a-i-1);
  else if(x >= 0) s = n-i-1;
  return s;
}

int index(int x) {
  int s = 0;
  rep(i,n-1) s += count(x, i);
  return s;
}

int solve(int k) {
  int l = -(int)1e18-1, r = (int)1e18+1;
  while(r - l > 1) {
    int m = (l + r) / 2;
    if(index(m) < k) l = m;
    else r = m;
  }
  return r;
}

signed main() {
  int k;
  cin >> n >> k;
  rep(i,n) cin >> a[i];
  sort(a, a+n);
  
  cout << solve(k) << endl;
  return 0;
}
```