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 azc {

	public static void main(String[] args) 
	{

		Scanner cin =new Scanner(System.in);
		int t=cin.nextInt();
		for(int i =0;i<t;i++)
		{
			int n=cin.nextInt();
			long arr[]= new long[n];
			for (int j = 0; j < arr.length; j++) 
			{
				arr[j]=cin.nextLong();
			}
			Arrays.sort(arr);
			if((n&1)==0)
			{
				int k=1;
				for (int j = n/2; j<arr.length; j++)
				{
					System.out.println(arr[j]);
					System.out.println(arr[j-k]);
					k=k+2;
				}
			}
			else
			{
				System.out.println(arr[n/2]);
				int k=2;
				for (int j = n/2+1; j<arr.length; j++)
				{
					System.out.println(arr[j]);
					System.out.println(arr[j-k]);
					k=k+2;
				}
			}
			
		}
	}

}