#include <algorithm>
#include <iostream>
#include <map>
#include <set>
#include <vector>
using namespace std;
int main() {
    long long N, K;
    cin >> N >> K;
    vector<long> A(N);
    for (int i; i < N; i++) {
        cin >> A[i];
    }
    sort(A.begin(), A.end());

    long long left = -2e18, right = 2e18;
    while (right - left > 1) {
        long long mid = (left + right) / 2, count = 0;
        for (int i = 0; i < N; i++) {
            int l = i, r = N;
            while (r - l > 1) {
                int m = (l + r) / 2;
                if (A[i] * A[m] <= mid)
                    (A[i] < 0) ? r = m : l = m;
                else
                    (A[i] < 0) ? l = m : r = m;
            }
            count += (A[i] < 0) ? N - r : l - i;
        }
        (count < K) ? left = mid : right = mid;
    }
    cout << right << "\n";
}

### 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 <algorithm>
#include <iostream>
#include <map>
#include <set>
#include <vector>
using namespace std;
int main() {
    long long N, K;
    cin >> N >> K;
    vector<long> A(N);
    for (int i; i < N; i++) {
        cin >> A[i];
    }
    sort(A.begin(), A.end());

    long long left = -2e18, right = 2e18;
    while (right - left > 1) {
        long long mid = (left + right) / 2, count = 0;
        for (int i = 0; i < N; i++) {
            int l = i, r = N;
            while (r - l > 1) {
                int m = (l + r) / 2;
                if (A[i] * A[m] <= mid)
                    (A[i] < 0) ? r = m : l = m;
                else
                    (A[i] < 0) ? l = m : r = m;
            }
            count += (A[i] < 0) ? N - r : l - i;
        }
        (count < K) ? left = mid : right = mid;
    }
    cout << right << "\n";
}

```