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:
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import java.util.*;
public class Main
{
	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		
		int t = sc.nextInt();
		
		for(int i =0;i<t;i++)
		{
		    int n = sc.nextInt();
		    Vector<Integer> vec = new Vector<>(n);
		    Vector<Integer> vec1 = new Vector<>(n);
		    
		    for(int j = 0;j<n;j++)
		        vec.add(sc.nextInt());
		  
		    Collections.sort(vec);
		    //System.out.println(vec);
		    int back = (n-1)/2,front = back + 1;
		    while(back>=0 || front!=n)
		    {
		        if(back>=0)
		        {
		            vec1.add(vec.get(back));
		            back--;
		        }
		        if(front!=n)
		        {
		            vec1.add(vec.get(front));
		            front++; 
		        }
		    }
		   for(int j = 0;j<n;j++)
		        System.out.print(vec1.get(j)+" ");
		    System.out.println();
		    
		}
		
		
	}
}