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:
/*
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Just live life young
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*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1339B
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         while(T-->0)
         {
            int N = Integer.parseInt(infile.readLine());
            int[] arr = new int[N];
            st = new StringTokenizer(infile.readLine());
            for(int i=0; i < N; i++)
               arr[i] = Integer.parseInt(st.nextToken());
            sort(arr);
            ArrayList<Integer> ls = new ArrayList<Integer>();
            int dex1 = N/2;
            int dex2 = dex1+1;
            if(N%2 == 0)
            {
               dex1--;
               dex2--;
            }
            for(int i=0; i < N; i++)
            {
               if(i%2 == 0)
               {
                  ls.add(arr[dex1]);
                  dex1--;
               }
               else
               {
                  ls.add(arr[dex2]);
                  dex2++;
               }
            }
            for(int x: ls)
               sb.append(x+" ");
            sb.append("\n");
         }
         System.out.print(sb);
      }
      public static void sort(int[] arr)
      {
         //stable heap sort
         PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
         for(int a: arr)
            pq.add(a);
         for(int i=0; i < arr.length; i++)
            arr[i] = pq.poll();
      }
   }