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 Expert {

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

    public static void main(String[] args) {
        int T = in.nextInt();
        while (T-- > 0) {
            ArrayList<Integer> list = new ArrayList<Integer>();
            int n = in.nextInt();
            int tmp = n;
            while (tmp-- > 0) {
                list.add(in.nextInt());
            }
            Collections.sort(list);
            int l = 0;
            int r = n - 1;
            ArrayList<Integer> ans = new ArrayList<Integer>();

            while (r > l) {

                ans.add(list.get(r--));
                ans.add(list.get(l++));
            }
            if (n % 2 != 0) {
                System.out.print(list.get(n/2)+" ");
            }
            for (int i = ans.size() - 1; i >= 0; i--) {
                System.out.print(ans.get(i) + " ");
            }
            System.out.println("");
        }
    }

}