#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const ll INF = ll(1e18)+1;
vector<ll> a;
ll n, k;
ll ok(ll x0, ll ai){
    ll r, l;
    ll ans;
    if(ai<0){
        l = -1;r = n;
        while(l+1<r){
            ll x = (r+l)/2;
            if(ai*a[x]<x0) r=x;
            else l=x;
        }
        ans = n-r;      
    }else{
        l = -1; r = n;
        while(l+1<r){
            ll x = (r+l)/2;
            if(ai*a[x]<x0) l = x;
            else r=x;
        }
        ans = r;
    }
    if(ai*ai<x0)ans--;
    return ans;
}

int main(){
    cin >> n >> k;
    a.resize(n);
    rep(i,n){
        cin >>  a[i];
    }
    sort(a.begin(), a.end());
    ll l=-INF, r=INF;
    while(l+1<r){
        ll x = (l+r)/2;
        ll tot = 0;
        rep(i,n){
            tot += ok(x, a[i]);
        }
        tot /= 2;
        if(tot < k) l=x;
        else r=x;
    }
    cout << l << 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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const ll INF = ll(1e18)+1;
vector<ll> a;
ll n, k;
ll ok(ll x0, ll ai){
    ll r, l;
    ll ans;
    if(ai<0){
        l = -1;r = n;
        while(l+1<r){
            ll x = (r+l)/2;
            if(ai*a[x]<x0) r=x;
            else l=x;
        }
        ans = n-r;      
    }else{
        l = -1; r = n;
        while(l+1<r){
            ll x = (r+l)/2;
            if(ai*a[x]<x0) l = x;
            else r=x;
        }
        ans = r;
    }
    if(ai*ai<x0)ans--;
    return ans;
}

int main(){
    cin >> n >> k;
    a.resize(n);
    rep(i,n){
        cin >>  a[i];
    }
    sort(a.begin(), a.end());
    ll l=-INF, r=INF;
    while(l+1<r){
        ll x = (l+r)/2;
        ll tot = 0;
        rep(i,n){
            tot += ok(x, a[i]);
        }
        tot /= 2;
        if(tot < k) l=x;
        else r=x;
    }
    cout << l << endl;
    return 0;
}
```