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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Main {
   public static void main(String[] args) {
//        Scanner sc = new Scanner(System.in);
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = sc.nextInt();
        outer:
        while (t-- >= 1) {
            int n=sc.nextInt();
            int a[]=sc.readArray(n);
            int ans[]= new int[n];
            sort(a);
            if (n%2==0){
                int k=n-1;
                for (int i=0;i<n/2;i++){
                    ans[k--]=a[n-1-i];
                    ans[k--]=a[i];
                }
            }
            else{
               ans[0]=a[n/2];
               int k=n-1;
               for (int i=0;i<n/2;i++){
                   ans[k--]=a[n-1-i];
                   ans[k--]=a[i];
               }
            }
            for (int i=0;i<n;i++){
                System.out.print(ans[i]+" ");
            }
            System.out.println();


            out.flush();
        }

        }

    static void sort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }
    static void sortReverse(long[] a) {
        ArrayList<Long> l=new ArrayList<>();
        for (long i:a) l.add(i);
//        Collections.sort.(l);
        Collections.sort(l,Collections.reverseOrder());
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long[] readArrayLong(long n) {
            long[] a=new long[(int)n];
            for (int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }


}