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.Arrays;
import java.util.Scanner;

public class Solution {
	public static int[] solve(int n, int[] arr) {
		if (n <= 2) {
			return arr;
		}
		int[] ans = new int[n];
		int temp = n - 1;
		for (int i = 0; i < n / 2; i++) {
			int first = arr[i];
			int last = arr[n - i - 1];
			ans[temp] = last;
			ans[temp - 1] = first;
			temp -= 2;
		}

		if (n % 2 != 0) {
			ans[0] = arr[n / 2];
		}

		return ans;
	}

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int t = s.nextInt();
		s.nextLine();
		for (int j = 0; j < t; j++) {
			int n = s.nextInt();
			int[] arr = new int[n];
			s.nextLine();
			for (int i = 0; i < n; i++) {
				arr[i] = s.nextInt();
			}
			Arrays.sort(arr);
			int[] ans = solve(n, arr);
			for (int k : ans) {
				System.out.print(k + " ");
			}
			System.out.println();
		}
	}
}