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.math.*;
import java.util.*;

public class Solve {
    FastScanner in;
    PrintWriter out;

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) {
            l.add(i);
        }
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) {
            a[i]=l.get(i);
        }
    }

    void solve() {
        int T = in.nextInt();
        while (0 < T--) {
            int n = in.nextInt();
            int[] a = in.nextIntArray(n);
            sort(a);

            int[] res = new int[n];
            int i = 0;
            int j = n-1;

            for (int x = 0; x < n; x++) {
                res[x] = (x % 2 == 0) ? a[j--] : a[i++];
            }

            for (int x = n-1; 0 <= x; x--) {
                out.print(res[x] + " ");
            }
            out.println();
        }
    }

    void run() {
        try {
            in = new FastScanner(new File("Solve.in"));
            out = new PrintWriter(new File("Solve.out"));

            solve();

            out.close();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }

    void runIO() {

        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);

        solve();

        out.close();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        public FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return null;
                st = new StringTokenizer(s);
            }
            return st.nextToken();
        }

        boolean hasMoreTokens() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return false;
                st = new StringTokenizer(s);
            }
            return true;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
            }
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextLong();
            }
            return a;
        }

        BigInteger nextBigInteger() {
            return new BigInteger(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        double[] nextDoubleArray(int n) {
            double[] a = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextDouble();
            }
            return a;
        }

    }

    public static void main(String[] args) {
        new Solve().runIO();
    }
}