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.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        StringBuilder sb=new StringBuilder();
        int t =sc.nextInt();
        while (t-->0){
            int n=sc.nextInt();
            Integer[] arr=new Integer[n];
            for(int i=0;i<n;i++)arr[i]=sc.nextInt();
            Arrays.sort(arr);
            int l=0,r=n-1;
            LinkedList<Integer> ll=new LinkedList<>();
            while(l<=r){
                if(l!=r){
                    ll.addFirst(arr[r]);
                    ll.addFirst(arr[l]);
                }
                else ll.addFirst(arr[l]);
                l++;
                r--;
            }
            for(int i:ll)sb.append(i+" ");
            sb.append("\n");
        }
        System.out.println(sb);
    }
}
class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); }
    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    int nextInt() { return Integer.parseInt(next()); }
    long nextLong() { return Long.parseLong(next()); }
    double nextDouble() { return Double.parseDouble(next()); }
    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}