#include <bits/stdc++.h>

#define rep(i,n) for (int i = 0; i < (n); ++i)

using namespace std;
typedef long long ll;
typedef pair<int, int> pint;

const int INF = 1 << 30;
const ll LINF = 1LL << 60;
const int MOD = 1e9 + 7;


int main() {
    ll n, k; cin >> n >> k;
    vector<ll> a(n); rep(i, n) cin >> a[i];
    sort(a.begin(), a.end());

    ll r = 1e18 + 1;
    ll l = -1e18 - 1;
    ll mid;
    while (r-l>1) {
        mid = (r+l) / 2;
        ll cnt = 0;
        rep(i, n) {
            if (a[i] > 0) {
                cnt += partition_point(a.begin(), a.begin()+i, [&](const ll v) {return v * a[i] <= mid;}) - a.begin();
            } else if (a[i] == 0) {
                if (mid >= 0) cnt += i;
            } else {
                cnt += a.begin()+i - partition_point(a.begin(), a.begin()+i, [&](const ll v) {return v * a[i] > mid;});
            }
        }
        if (cnt < k) l = mid;
        else r = mid;
    }
    
    cout << r << endl;



    return 0;
}


    

### Prompt

Construct a Cpp code solution to the problem outlined:

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>

#define rep(i,n) for (int i = 0; i < (n); ++i)

using namespace std;
typedef long long ll;
typedef pair<int, int> pint;

const int INF = 1 << 30;
const ll LINF = 1LL << 60;
const int MOD = 1e9 + 7;


int main() {
    ll n, k; cin >> n >> k;
    vector<ll> a(n); rep(i, n) cin >> a[i];
    sort(a.begin(), a.end());

    ll r = 1e18 + 1;
    ll l = -1e18 - 1;
    ll mid;
    while (r-l>1) {
        mid = (r+l) / 2;
        ll cnt = 0;
        rep(i, n) {
            if (a[i] > 0) {
                cnt += partition_point(a.begin(), a.begin()+i, [&](const ll v) {return v * a[i] <= mid;}) - a.begin();
            } else if (a[i] == 0) {
                if (mid >= 0) cnt += i;
            } else {
                cnt += a.begin()+i - partition_point(a.begin(), a.begin()+i, [&](const ll v) {return v * a[i] > mid;});
            }
        }
        if (cnt < k) l = mid;
        else r = mid;
    }
    
    cout << r << endl;



    return 0;
}


    

```