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.*;
import java.math.*;

public class B {

	public void realMain() throws Exception {

		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in), 1000000);

		String in = fin.readLine();

		String[] ar = in.split(" ");

		int T = Integer.parseInt(ar[0]);
	for(int t = 0; t < T; t++) {
		
			int ret = 0;
		            int c = fin.read(); 
		            while (c <= ' ') 
                		c = fin.read(); 
		            boolean neg = (c == '-'); 
		            if (neg) 
		                c = fin.read(); 
		            do
		            { 
		                ret = ret * 10 + c - '0'; 
		            }  while ((c = fin.read()) >= '0' && c <= '9'); 
  
  		          if (neg) 
                		ret *= -1;

			int n = ret;

		Integer[] s = new Integer[n];

		for(int i = 0; i < n; i++) {
			
				ret = 0;
		            c = fin.read(); 
		            while (c <= ' ') 
                		c = fin.read(); 
		            neg = (c == '-'); 
		            if (neg) 
		                c = fin.read(); 
		            do
		            { 
		                ret = ret * 10 + c - '0'; 
		            }  while ((c = fin.read()) >= '0' && c <= '9'); 
  
  		          if (neg) 
                		ret *= -1;

			s[i] = ret;


		}

		Arrays.sort(s);

		StringBuilder sb = new StringBuilder("");

		if(n % 2 == 1) {
			sb.append( s[n / 2] );
			sb.append(' ');

			sb.append( s[n / 2 - 1]);
			sb.append(' ');	

			sb.append( s[n / 2 + 1]);
			sb.append(' ');

			int goingdown = n / 2 - 2;
			int goingup = n / 2 + 2;

			boolean booldown = true;

			int numdone = 3;

			while(numdone < n) {
				if(booldown) {
					sb.append( s[goingdown] );
					sb.append(' ');
	
					goingdown--;
				} else {
					sb.append( s[goingup] );
					sb.append(' ');

					goingup++;
				}

				booldown = !booldown;

				numdone++;
			}

		} else {
			sb.append( s[n / 2 - 1] );
			sb.append(' ');

			sb.append( s[n / 2] );
			sb.append(' ');

			int goingdown = n / 2 - 2;
			int goingup = n / 2 + 1;

			int numdone = 2;

			boolean booldown = true;

			while(numdone < n) {
				if(booldown) {
					sb.append( s[goingdown] );
					sb.append(' ');
	
					goingdown--;
				} else {
					sb.append( s[goingup] );
					sb.append(' ');

					goingup++;
				}

				booldown = !booldown;

				numdone++;
			}
		}

		System.out.println(sb);

	}
		



	}


	public static void main(String[] args) throws Exception {
		B b = new B();
		b.realMain();
	}
}