#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
 
 
int main() {
    ll n,k;
    cin >> n >> k;
    vector<ll> a(n);
    for (ll i=0;i<n;i++) cin >> a[i];
    sort(a.begin(),a.end());
    ll right=ll(1e18)+1;
    ll left=-right;
    while (right-left>1) {
        ll mid=(left+right)/2;
        ll cnt=0;
        for (ll i=0;i<n;i++) {
            if (a[i]<0) {
                ll ng=-1,ok=n;
                while (ok-ng>1) {
                    ll m=(ng+ok)/2;
                    if (a[i]*a[m]<mid) ok=m;
                    else ng=m;
                }
                cnt+=n-ok;
            } else {
                ll ng=-1,ok=n;
                while (ok-ng>1) {
                    ll m=(ng+ok)/2;
                    if (a[i]*a[m]<mid) ng=m;
                    else ok=m;
                }
                cnt+=ok;
            }
            if (a[i]*a[i]<mid) cnt--;
        }
        if (cnt/2<k) left=mid;
        else right=mid;
    }
    cout << left << endl;
}
### Prompt

Please formulate 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>
typedef long long ll;
using namespace std;
 
 
int main() {
    ll n,k;
    cin >> n >> k;
    vector<ll> a(n);
    for (ll i=0;i<n;i++) cin >> a[i];
    sort(a.begin(),a.end());
    ll right=ll(1e18)+1;
    ll left=-right;
    while (right-left>1) {
        ll mid=(left+right)/2;
        ll cnt=0;
        for (ll i=0;i<n;i++) {
            if (a[i]<0) {
                ll ng=-1,ok=n;
                while (ok-ng>1) {
                    ll m=(ng+ok)/2;
                    if (a[i]*a[m]<mid) ok=m;
                    else ng=m;
                }
                cnt+=n-ok;
            } else {
                ll ng=-1,ok=n;
                while (ok-ng>1) {
                    ll m=(ng+ok)/2;
                    if (a[i]*a[m]<mid) ng=m;
                    else ok=m;
                }
                cnt+=ok;
            }
            if (a[i]*a[i]<mid) cnt--;
        }
        if (cnt/2<k) left=mid;
        else right=mid;
    }
    cout << left << endl;
}
```