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.*;
public class Sample
{
   static void merge(int arr[], int l, int m, int r)
    {
        int n1 = m - l + 1;
        int n2 = r - m;
        int L[] = new int [n1];
        int R[] = new int [n2];
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
        int i = 0, j = 0;
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
    static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            int m = (l+r)/2;
            sort(arr, l, m);
            sort(arr , m+1, r);
            merge(arr, l, m, r);
        }
    }
  public static void main(String[] args)
  {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    while(t-->0)
    {
      int n = in.nextInt();
      int[] are = new int[n];
      for(int i=0; i<n; i++)
        are[i]=in.nextInt();
      sort(are, 0, n-1);
      int mid;
      if(n%2==0)
        mid=n/2-1;
      else
        mid=n/2;
      System.out.print(are[mid]+" ");
      for(int i=1; i<=mid; i++)
      {
        System.out.print(are[mid+i]+" "+are[mid-i]+" ");
      }
      if(n%2==0)
        System.out.print(are[n-1]);
      System.out.println();
    }
  }
}