#include<bits/stdc++.h>
using namespace std;
#define int long long
constexpr int mu = 1e9;
signed main(){
    int a[200000],n,k;
    cin>>n>>k;
    for(int i=0;i<n;i++) cin>>a[i];
    sort(a,a+n);
    int lb = -(mu*mu) ,ub =mu*mu;// 二分区间范围。
    while(ub - lb>1){
        int mid = (lb+ub)/2,cnt=0;//cnt 用来统计对的数量。
        for(int i=0;i<n;i++){
            if(a[i] < 0){
                cnt +=n- (lower_bound(a,a+n,(long double )mid/a[i])-a);
            }
            else if( a[i] >0){
                cnt += upper_bound(a,a+n,(long double)mid/a[i])-a;
            }
            else if( mid >= 0){
                // a[i]为0 且 任意值x >=0的话就有n对。
                cnt+=n;
            }
            // 去掉本身重复的。
            if( a[i] * a[i] <= mid) cnt--;
        }

        //判定条件，注意cnt是对数;
        (cnt>=2*k?ub:lb) = mid;
    }
    cout<<ub<<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>
using namespace std;
#define int long long
constexpr int mu = 1e9;
signed main(){
    int a[200000],n,k;
    cin>>n>>k;
    for(int i=0;i<n;i++) cin>>a[i];
    sort(a,a+n);
    int lb = -(mu*mu) ,ub =mu*mu;// 二分区间范围。
    while(ub - lb>1){
        int mid = (lb+ub)/2,cnt=0;//cnt 用来统计对的数量。
        for(int i=0;i<n;i++){
            if(a[i] < 0){
                cnt +=n- (lower_bound(a,a+n,(long double )mid/a[i])-a);
            }
            else if( a[i] >0){
                cnt += upper_bound(a,a+n,(long double)mid/a[i])-a;
            }
            else if( mid >= 0){
                // a[i]为0 且 任意值x >=0的话就有n对。
                cnt+=n;
            }
            // 去掉本身重复的。
            if( a[i] * a[i] <= mid) cnt--;
        }

        //判定条件，注意cnt是对数;
        (cnt>=2*k?ub:lb) = mid;
    }
    cout<<ub<<endl;
}
```