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 P1339B {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while(t > 0) {
            int n = in.nextInt();
            int[] arr = new int[n];
            int[] temp = new int[n];

            for(int i = 0; i < n; i++) {
                arr[i] = in.nextInt();
            }
            Arrays.sort(arr);

            for(int i = 0; i < n / 2;  i++) {
                temp[n  - 2 * i - 1] = arr[i];
            }

            if(n % 2 == 1) {
                temp[0] = arr[n / 2];
                for(int i = n / 2 + 1; i < n; i++) {
                    temp[2 * i - n] = arr[i];
                }
            }
            else {
                for(int i = n / 2; i < n; i++) {
                    temp[2 * i - n] = arr[i];
                }
            }

            for(int i  = 0; i < n - 1; i++) {
                System.out.print(temp[i] + " ");
            }
            System.out.println(temp[n - 1]);
            t--;
        }
    }
}