#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

const ll mod = 1e9 + 7;

int main() {
    ll n, k, a[200000];
    cin >> n >> k;
    for (ll i = 0; i < n; i++)
        cin >> a[i];
    sort(a, a + n);
    ll ub = mod * mod, lb = -(mod * mod);
    while (ub - lb > 1) {
        ll mid = (ub + lb) / 2, cnt = 0;
        for (ll i = 0; i < n; i++) {
            if (a[i] > 0)
                cnt += upper_bound(a, a + n, (long double)mid / a[i]) - a;
            else if (a[i] < 0)
                cnt += n - (lower_bound(a, a + n, (long double)mid / a[i]) - a);
            else if (mid >= 0)
                cnt += n;
            if (a[i] * a[i] <= mid)
                cnt--;
        }
        (cnt >= 2 * k ? ub : lb) = mid;
    }
    cout << ub << endl;
}

### Prompt

Generate a cpp solution to 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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (n); i++)

const ll mod = 1e9 + 7;

int main() {
    ll n, k, a[200000];
    cin >> n >> k;
    for (ll i = 0; i < n; i++)
        cin >> a[i];
    sort(a, a + n);
    ll ub = mod * mod, lb = -(mod * mod);
    while (ub - lb > 1) {
        ll mid = (ub + lb) / 2, cnt = 0;
        for (ll i = 0; i < n; i++) {
            if (a[i] > 0)
                cnt += upper_bound(a, a + n, (long double)mid / a[i]) - a;
            else if (a[i] < 0)
                cnt += n - (lower_bound(a, a + n, (long double)mid / a[i]) - a);
            else if (mid >= 0)
                cnt += n;
            if (a[i] * a[i] <= mid)
                cnt--;
        }
        (cnt >= 2 * k ? ub : lb) = mid;
    }
    cout << ub << endl;
}

```