#include <bits/stdc++.h>
using namespace std;
using ll  = long long;
using vl  = vector<ll>;

#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rng(i,a,b) for(int i = (a); i < (b); ++i)
#define all(a) (a).begin(), (a).end()

int main() {
    ll N, K; cin >> N >> K;
    vl A(N);
    rep(i,N) cin >> A[i];
    sort(all(A));

    ll ng = 2e18, ok = -ng;
    while (ng - ok > 1) {
        ll x = (ok + ng) / 2, cnt = 0;
        rep(i,N) {
            if (A[i] < 0) cnt += N - (upper_bound(all(A), (long double)x/A[i]) - A.begin());
            if (A[i] > 0) cnt += lower_bound(all(A), (long double)x/A[i]) - A.begin();
            if (A[i] == 0 and x > 0) cnt += N;
            if (A[i]*A[i] < x) cnt--;
        }
        (cnt/2 < K ? ok : ng) = x;
    }
    cout << ok << endl;
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using vl  = vector<ll>;

#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rng(i,a,b) for(int i = (a); i < (b); ++i)
#define all(a) (a).begin(), (a).end()

int main() {
    ll N, K; cin >> N >> K;
    vl A(N);
    rep(i,N) cin >> A[i];
    sort(all(A));

    ll ng = 2e18, ok = -ng;
    while (ng - ok > 1) {
        ll x = (ok + ng) / 2, cnt = 0;
        rep(i,N) {
            if (A[i] < 0) cnt += N - (upper_bound(all(A), (long double)x/A[i]) - A.begin());
            if (A[i] > 0) cnt += lower_bound(all(A), (long double)x/A[i]) - A.begin();
            if (A[i] == 0 and x > 0) cnt += N;
            if (A[i]*A[i] < x) cnt--;
        }
        (cnt/2 < K ? ok : ng) = x;
    }
    cout << ok << endl;
    return 0;
}

```