Problem: You have array of n numbers a_{1}, a_{2}, …, a_{n}. 

Rearrange these numbers to satisfy |a_{1} - a_{2}| ≤ |a_{2} - a_{3}| ≤ … ≤ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.

Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.

You have to answer independent t test cases.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^{4}) — the number of test cases.

The first line of each test case contains single integer n (3 ≤ n ≤ 10^{5}) — the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.

The second line of each test case contains n integers a_{1}, a_{2}, …, a_{n} (-10^{9} ≤ a_{i} ≤ 10^{9}).

Output

For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.

Example

Input


2
6
5 -2 4 8 6 5
4
8 1 4 2


Output


5 5 4 6 8 -2
1 2 4 8

Note

In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 ≤ |a_{2} - a_{3}| = 1 ≤ |a_{3} - a_{4}| = 2 ≤ |a_{4} - a_{5}| = 2 ≤ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".

In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 ≤ |a_{2} - a_{3}| = 2 ≤ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {

     static Scanner i = new Scanner(System.in);

    static public void main(String[] args) throws IOException {
        int t=i.nextInt();
        for(int j=0;j<t;j++){
            int n = i.nextInt();
            int a[]=new int [n];
            for (int k = 0;k < n; k++){
                a[k]=i.nextInt();
            }
            Arrays.sort(a);
            int v[]=new int [n];
            int m1,m2;
//            System.out.println(n);
            if(n%2==0){
                m2=n/2;
                m1=m2-1;
            }
            else{
                m1=n/2;
                m2=m1+1;
            }
//            System.out.println(m1+" "+m2);
            for (int k = 0;k < n; k++){
                if(k%2==0){
                    v[k]=a[m1];
                    m1--;
                }
                else{
                    v[k]=a[m2];
                    m2++;
                }
            }
            for (int k = 0;k < n; k++){
                System.out.print(v[k]+" ");
            }
            System.out.println();
            }
    }
}