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.Scanner;
import java.util.Arrays;

public class mySolution {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int numCases = scan.nextInt();
        for (int c = 0; c < numCases; c++) {
            int numVals = scan.nextInt();
            int[] val = new int[numVals];
            for (int v = 0; v < numVals; v++) {
                val[v] = scan.nextInt();
            }
            Arrays.sort(val);

            // Arbitrarily pick a starting side
            boolean leftTurn = false;
            int lp = 0;
            int rp = numVals-1;
            int[] rearranged = new int[numVals];
            // Fill rearranged array backwards
            for (int np = numVals-1; np >= 0; np--) {
                if (leftTurn) {
                    rearranged[np] = val[lp];
                    lp++;
                } else {
                    rearranged[np] = val[rp];
                    rp--;
                }
                leftTurn = !leftTurn;
            }

            // Print result
            for (int i = 0; i < numVals; i++) {
                System.out.print(rearranged[i]);
                if (i < numVals - 1) {
                    System.out.print(" ");
                }
            }
            System.out.println();
        }
    }
}