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 P3 {
	static long mod=(int)1e9+7;
	static int n;
	static  long a[]=new long[5005];
static 	long b[]=new long[5005];
	static long dp[][]=new long[5005][5005];
	public static void main(String[] args)
	{
		Scanner sc=new Scanner(System.in);    
		int t=sc.nextInt();
		 while(t-->0) {
	            int n = sc.nextInt();
	            int a[]= new int[n];
	           for(int i=0;i<n;i++) {
	        	   a[i]=sc.nextInt();
	           }
	           Arrays.sort(a);
	           if((n&1)==0) {
	        	  
	        	   for(int i=(n-1)/2;i>=0;i--) {
	        		   System.out.print(a[i]+" ");
	        		   System.out.print(a[n-1-i]+" ");
	        	   }
	           }else {
	        	   System.out.print(a[n/2]+" ");
	        	for(int i=n/2-1;i>=0;i--) {
	        		System.out.print(a[i]+" ");
	        		   System.out.print(a[n-1-i]+" ");
	        		
	        	}
	           }
	           System.out.println();
	        }
	}
}