#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>

using ll = long long;

using namespace std;
#define MAX (ll)(1e18)

int main() {
    ll n, k;
    cin >> n >> k;
    ll a[n];
    for (auto &ai:a)
        cin >> ai;
    sort(a, a+n);
    ll l = -MAX, r = MAX;
    while (r - l > 1) {
        ll m = (l + r) / 2,
           sum = 0;
        for (int i=0; i<n; i++) {
            if (a[i] >= 0) {
                auto pp = partition_point(a+i+1, a+n, [&](ll x) { return x * a[i] <= m; });
                sum += pp - (a+i+1);
            }
            else {
                auto pp = partition_point(a+i+1, a+n, [&](ll x) { return x * a[i] > m; });
                sum += (a+n) - pp;
            }
        }
        if (sum < k) l = m;
        else r = m;
    }
    cout << r << '\n';
    return 0;
}
### Prompt

Your challenge is to write 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 <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>

using ll = long long;

using namespace std;
#define MAX (ll)(1e18)

int main() {
    ll n, k;
    cin >> n >> k;
    ll a[n];
    for (auto &ai:a)
        cin >> ai;
    sort(a, a+n);
    ll l = -MAX, r = MAX;
    while (r - l > 1) {
        ll m = (l + r) / 2,
           sum = 0;
        for (int i=0; i<n; i++) {
            if (a[i] >= 0) {
                auto pp = partition_point(a+i+1, a+n, [&](ll x) { return x * a[i] <= m; });
                sum += pp - (a+i+1);
            }
            else {
                auto pp = partition_point(a+i+1, a+n, [&](ll x) { return x * a[i] > m; });
                sum += (a+n) - pp;
            }
        }
        if (sum < k) l = m;
        else r = m;
    }
    cout << r << '\n';
    return 0;
}
```