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.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class b633 {
  public static class FastScanner {
    private BufferedReader br;
    private StringTokenizer st;
    public FastScanner(InputStream inputStream){
      br = new BufferedReader(new InputStreamReader(inputStream));
    }
    public String nextLine() throws IOException {
      return br.readLine();
    }
    public String next() {
      while (st == null || !st.hasMoreTokens()){
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException ignored){}
      }
      return st.nextToken();
    }
    public int nextInt() {
      return Integer.parseInt(next());
    }
    public long nextLong() {
      return Long.parseLong(next());
    }

  }
  public static void main(String[] args) {
    FastScanner in = new FastScanner(System.in);
    int t = in.nextInt();
    while (t -- > 0) {
      int n = in.nextInt();
      ArrayList<Integer> ar = new ArrayList<Integer>();
      for (int i = 0; i < n; i++) ar.add(in.nextInt());
      Collections.sort(ar);
      int left = 0;
      int right = n - 1;
      int turn = 0;
      int ans[] = new int[n];
      int idx = 0;
      while (left <= right) {

        if (turn == 0) {
          ans[idx++] = ar.get(left++);
        } else {
          ans[idx++] = ar.get(right--);
        }
        turn ^= 1;
      }
      for (int i = n-1; i >=  0; i--) {
        if (i != n-1) System.out.print(" ");
        System.out.print(ans[i]);
      }
      System.out.println();
    }
  }
}