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 SortedAdjacentDifferences {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			Integer arr[]=new Integer[n];
			Integer dec[]=new Integer[n];
			Integer inc[]=new Integer[n];
			for(int i=0;i<n;i++) {
				arr[i]=sc.nextInt();
				dec[i]=arr[i];
				inc[i]=arr[i];
			}
			
			Arrays.sort(dec,Collections.reverseOrder());
			Arrays.sort(inc);
			
			ArrayList<Integer> a=new ArrayList<Integer>();
			if(n%2==0) {
				for(int i=0;i<n/2;i++) {
					a.add(dec[i]);
					a.add(inc[i]);
				}
			}else {
				for(int i=0;i<Math.ceil((double)n/(double)2);i++) {
					if(i==Math.ceil((double)n/(double)2)-1) {
						a.add(dec[i]);
					}else {
						a.add(dec[i]);
						a.add(inc[i]);
					}
				}
			}
			
			for(int i=a.size()-1;i>=0;i--) {
				System.out.print(a.get(i)+" ");
			}
			System.out.println();
			
		}

	}

}