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.*;
import java.util.*;
public class b633 {

	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
        int t = scn.nextInt();
        while(t!=0)
        {
        	int n = scn.nextInt();
        	long arr[]=new long[n];
        	long ans[]=new long[n];
        	for(int i =0;i<n;i++)
        	{
        		arr[i]=scn.nextLong();
        	}
        	Arrays.sort(arr);
        	int i =0;
        	int j =n-1;
        	int k =0;
        	for( k =n-1;k>=0;k--)
        	{
        		ans[k]=arr[i];
        		i++;
        		k--;
        		ans[k]=arr[j];
        		j--;
        		if(i==j)
        		{
        			k--;
        			ans[k]=arr[i];
        			break;
        		}
        			
        	}
        	for(int f=0;f<n;f++)
        	{
        		System.out.print(ans[f]);
        		System.out.print(" ");
        	}
        	
        	System.out.println();
        	
        	t--;
        }

	}

}