#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
ll N, K; vector<ll> Ap, Am, Az;
ll getCount(ll x) {
  ll cnt = 0;
  for(ll m = 0, p = Ap.size()-1; m < Am.size(); ++m) {
    while(p >= 0 && x >= -Am[m]*Ap[p]) --p;
    cnt += Ap.size()-p-1;
  }
  for(ll z = 1; x >= 0 && z <= Az.size(); ++z) {
    cnt += N-z;
  }
  for(ll l = 0, r = Ap.size()-1; l < r; ++l) {
    while(l <= r && x < Ap[l]*Ap[r]) --r;
    if(l < r) cnt += r-l;
  }
  for(ll l = 0, r = Am.size()-1; l < r; ++l) {
    while(l <= r && x < Am[l]*Am[r]) --r;
    if(l < r) cnt += r-l;
  }
  return cnt;
}
int main() {
  scanf("%ld %ld", &N, &K);
  for(ll i = 1, A; i <= N; ++i) {
    scanf("%ld", &A);
    if(A > 0) Ap.push_back(A);
    else if(A < 0) Am.push_back(-A);
    else Az.push_back(A);
  }
  sort(Ap.begin(), Ap.end());
  sort(Am.begin(), Am.end());
  ll l = -1e18, r = 1e18;
  while(r-l > 1) {
    ll m = (l+r)/2;
    if(getCount(m) >= K) r = m;
    else l = m;
  }
  printf("%ld\n", r);
}
### 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 <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
ll N, K; vector<ll> Ap, Am, Az;
ll getCount(ll x) {
  ll cnt = 0;
  for(ll m = 0, p = Ap.size()-1; m < Am.size(); ++m) {
    while(p >= 0 && x >= -Am[m]*Ap[p]) --p;
    cnt += Ap.size()-p-1;
  }
  for(ll z = 1; x >= 0 && z <= Az.size(); ++z) {
    cnt += N-z;
  }
  for(ll l = 0, r = Ap.size()-1; l < r; ++l) {
    while(l <= r && x < Ap[l]*Ap[r]) --r;
    if(l < r) cnt += r-l;
  }
  for(ll l = 0, r = Am.size()-1; l < r; ++l) {
    while(l <= r && x < Am[l]*Am[r]) --r;
    if(l < r) cnt += r-l;
  }
  return cnt;
}
int main() {
  scanf("%ld %ld", &N, &K);
  for(ll i = 1, A; i <= N; ++i) {
    scanf("%ld", &A);
    if(A > 0) Ap.push_back(A);
    else if(A < 0) Am.push_back(-A);
    else Az.push_back(A);
  }
  sort(Ap.begin(), Ap.end());
  sort(Am.begin(), Am.end());
  ll l = -1e18, r = 1e18;
  while(r-l > 1) {
    ll m = (l+r)/2;
    if(getCount(m) >= K) r = m;
    else l = m;
  }
  printf("%ld\n", r);
}
```