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.IOException;
import java.io.InputStream;
import java.util.Arrays;


public class Solution {
    public static void main(String args[]) throws IOException {
        FastReader in = new FastReader(System.in);
        StringBuilder sb = new StringBuilder();
        int i,j;
        int t=in.nextInt();
        while(t-->0){
            int n=in.nextInt();
            Integer arr[]=new Integer[n];
            for(i=0;i<n;i++)
                arr[i]=in.nextInt();
            Arrays.sort(arr);
            int mid=n/2;
            i=mid-1;
            j=mid+1;
            sb.append(arr[mid]).append(" ");
            while(i>=0){
                sb.append(arr[i]).append(" ");
                i--;
                if(j>=n)
                    break;
                sb.append(arr[j]).append(" ");
                j++;
            }

            sb.append("\n");
        }
        System.out.print(sb);


    }
}

class FastReader {

    byte[] buf = new byte[2048];
    int index, total;
    InputStream in;

    FastReader(InputStream is) {
        in = is;
    }

    int scan() throws IOException {
        if (index >= total) {
            index = 0;
            total = in.read(buf);
            if (total <= 0) {
                return -1;
            }
        }
        return buf[index++];
    }

    String next() throws IOException {
        int c;
        for (c = scan(); c <= 32; c = scan()) ;
        StringBuilder sb = new StringBuilder();
        for (; c > 32; c = scan()) {
            sb.append((char) c);
        }
        return sb.toString();
    }

    int nextInt() throws IOException {
        int c, val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+') {
            c = scan();
        }
        for (; c >= '0' && c <= '9'; c = scan()) {
            val = (val << 3) + (val << 1) + (c & 15);
        }
        return neg ? -val : val;
    }

    long nextLong() throws IOException {
        int c;
        long val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+') {
            c = scan();
        }
        for (; c >= '0' && c <= '9'; c = scan()) {
            val = (val << 3) + (val << 1) + (c & 15);
        }
        return neg ? -val : val;
    }
}