#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const ll INF = 1ll << 61;

ll N, K;
vector<ll> A;
ll fdiv(ll x, ll y) { return x >= 0ll ? x / y : -((-x + y - 1) / y); }
ll count(ll x) {
  ll cnt = 0;
  // x 以下になるペアの個数
  for (ll i = 0; i < N; ++i) {
    if (A[i] == 0) {
      if (x >= 0) cnt += i;
    } else if (A[i] > 0) {
      ll f = fdiv(x, A[i]); // floor(x / A[i])
      ll j = upper_bound(A.begin(), A.end(), f) - A.begin();
      cnt += min(j, i);
    } else {
      ll c = -fdiv(x, (-A[i])); // ceil(x / A[i]) = - floor(x / -A[i])
      ll k = lower_bound(A.begin(), A.end(), c) - A.begin();
      cnt += max(0ll, i - k);
    }
  }
  return cnt;
}
int main() {
  cin >> N >> K;
  for (ll i = 0; i < N; ++i) {
    ll a;
    cin >> a;
    A.push_back(a);
  }
  sort(A.begin(), A.end());
  ll ok = INF, ng = -INF;
  while (ok - ng > 1) {
    ll mid = (ok + ng) / 2;
    if (count(mid) >= K)
      ok = mid;
    else
      ng = mid;
  }
  cout << ok << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 <bits/stdc++.h>
using namespace std;

typedef long long ll;

const ll INF = 1ll << 61;

ll N, K;
vector<ll> A;
ll fdiv(ll x, ll y) { return x >= 0ll ? x / y : -((-x + y - 1) / y); }
ll count(ll x) {
  ll cnt = 0;
  // x 以下になるペアの個数
  for (ll i = 0; i < N; ++i) {
    if (A[i] == 0) {
      if (x >= 0) cnt += i;
    } else if (A[i] > 0) {
      ll f = fdiv(x, A[i]); // floor(x / A[i])
      ll j = upper_bound(A.begin(), A.end(), f) - A.begin();
      cnt += min(j, i);
    } else {
      ll c = -fdiv(x, (-A[i])); // ceil(x / A[i]) = - floor(x / -A[i])
      ll k = lower_bound(A.begin(), A.end(), c) - A.begin();
      cnt += max(0ll, i - k);
    }
  }
  return cnt;
}
int main() {
  cin >> N >> K;
  for (ll i = 0; i < N; ++i) {
    ll a;
    cin >> a;
    A.push_back(a);
  }
  sort(A.begin(), A.end());
  ll ok = INF, ng = -INF;
  while (ok - ng > 1) {
    ll mid = (ok + ng) / 2;
    if (count(mid) >= K)
      ok = mid;
    else
      ng = mid;
  }
  cout << ok << endl;
  return 0;
}

```