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

const int64_t INF = 1LL << 60;

int N;
int64_t K;
vector<int64_t> A;

int solve(int i, int64_t x) {
    int l = i, r = N;
    while(r - l > 1) {
        int c = (l + r) / 2;
        (A[i] * A[c] <= x) ^ (A[i] > 0) ? r = c : l = c;
    }
    return A[i] > 0 ? l - i : N - r;
}

int main() {
    cin >> N >> K;
    A.resize(N);
    for(auto &e : A) cin >> e;
    sort(A.begin(), A.end());
    auto OK = [&](int64_t x) {
        int64_t cnt = 0;
        for(int i = 0; i < N; ++i) {
            if(A[i] == 0) cnt += (x >= 0 ? N - i - 1 : 0);
            else cnt += solve(i, x);
        }
        return cnt < K;
    };
    int64_t l = -INF, r = INF;
    while(r - l > 1) {
        int64_t c = (l + r) / 2;
        OK(c) ? l = c : r = c;
    }
    cout << r << '\n';
    return 0;
}

### 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;

const int64_t INF = 1LL << 60;

int N;
int64_t K;
vector<int64_t> A;

int solve(int i, int64_t x) {
    int l = i, r = N;
    while(r - l > 1) {
        int c = (l + r) / 2;
        (A[i] * A[c] <= x) ^ (A[i] > 0) ? r = c : l = c;
    }
    return A[i] > 0 ? l - i : N - r;
}

int main() {
    cin >> N >> K;
    A.resize(N);
    for(auto &e : A) cin >> e;
    sort(A.begin(), A.end());
    auto OK = [&](int64_t x) {
        int64_t cnt = 0;
        for(int i = 0; i < N; ++i) {
            if(A[i] == 0) cnt += (x >= 0 ? N - i - 1 : 0);
            else cnt += solve(i, x);
        }
        return cnt < K;
    };
    int64_t l = -INF, r = INF;
    while(r - l > 1) {
        int64_t c = (l + r) / 2;
        OK(c) ? l = c : r = c;
    }
    cout << r << '\n';
    return 0;
}

```