#include <bits/stdc++.h>
#define rep(i, n) for(ll i = 0; i < n; i++)
#define repr(i, n) for(ll i = n; i >= 0; i--)
#define reprr(i, m, n) for (ll i = m; i < n; i++)
#define inf LLONG_MAX
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;


int main()
{
    ll n, k; cin >> n >> k;
    vll a(n); rep(i, n) cin >> a[i];
    sort(all(a));

    ll l = -inf, r = inf;
    while (l + 1 < r) {
        ll cnt = 0;
        ll M = (l + r) / 2;

        rep(i, n) {
            ll l = i, r = n;
            while (l + 1 < r) {
                ll m = (l + r) / 2;
                if (a[i] * a[m] <= M) {
                    a[i] > 0 ? l = m : r = m;
                } else {
                    a[i] > 0 ? r = m : l = m;
                }
            }
            if (a[i] > 0) cnt += l - i;
            else cnt += n - r;
        }

        (cnt < k) ? l = M : r = M;
    }
    cout << r << endl;
    return 0;
}
### 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 <bits/stdc++.h>
#define rep(i, n) for(ll i = 0; i < n; i++)
#define repr(i, n) for(ll i = n; i >= 0; i--)
#define reprr(i, m, n) for (ll i = m; i < n; i++)
#define inf LLONG_MAX
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;


int main()
{
    ll n, k; cin >> n >> k;
    vll a(n); rep(i, n) cin >> a[i];
    sort(all(a));

    ll l = -inf, r = inf;
    while (l + 1 < r) {
        ll cnt = 0;
        ll M = (l + r) / 2;

        rep(i, n) {
            ll l = i, r = n;
            while (l + 1 < r) {
                ll m = (l + r) / 2;
                if (a[i] * a[m] <= M) {
                    a[i] > 0 ? l = m : r = m;
                } else {
                    a[i] > 0 ? r = m : l = m;
                }
            }
            if (a[i] > 0) cnt += l - i;
            else cnt += n - r;
        }

        (cnt < k) ? l = M : r = M;
    }
    cout << r << endl;
    return 0;
}
```