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

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N,K; cin >> N >> K;
    vector<ll> A(N);
    for (int i=0;i<N;++i) cin >> A[i];
    sort(A.begin(),A.end());
    ll ub=1e18,lb=-ub; ++ub,--lb;
    while(ub-lb>1){
        ll mid=(ub+lb)>>1LL,cnt=0;
        for (int i=0;i<N;++i){
            if (A[i]==0){
                if (0<=mid) cnt+=N-1;
                continue;
            }
            if (A[i]>0) cnt+=upper_bound(A.begin(),A.end(),mid/A[i]+(mid<0&&mid%A[i]!=0?-1:0))-A.begin();
            else cnt+=A.end()-lower_bound(A.begin(),A.end(),mid/A[i]+(mid<0&&mid%A[i]!=0?1:0));
            if (A[i]*A[i]<=mid) --cnt;
        }
        cnt/=2;
        (K<=cnt?ub:lb)=mid;
    }
    cout << ub << endl;
}

### 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;
typedef long long ll;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N,K; cin >> N >> K;
    vector<ll> A(N);
    for (int i=0;i<N;++i) cin >> A[i];
    sort(A.begin(),A.end());
    ll ub=1e18,lb=-ub; ++ub,--lb;
    while(ub-lb>1){
        ll mid=(ub+lb)>>1LL,cnt=0;
        for (int i=0;i<N;++i){
            if (A[i]==0){
                if (0<=mid) cnt+=N-1;
                continue;
            }
            if (A[i]>0) cnt+=upper_bound(A.begin(),A.end(),mid/A[i]+(mid<0&&mid%A[i]!=0?-1:0))-A.begin();
            else cnt+=A.end()-lower_bound(A.begin(),A.end(),mid/A[i]+(mid<0&&mid%A[i]!=0?1:0));
            if (A[i]*A[i]<=mid) --cnt;
        }
        cnt/=2;
        (K<=cnt?ub:lb)=mid;
    }
    cout << ub << endl;
}

```