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.*;

public class B {

	public static void main(String[] args) {

		Scanner nik = new Scanner(System.in);
		int t = nik.nextInt();
		StringBuilder sb = new StringBuilder();

		while (t-- > 0) {
			int n = nik.nextInt();
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = nik.nextInt();
			}
			Arrays.sort(a);

			int l = 0;
			int r = n - 1;
			StringBuilder st = new StringBuilder();
			ArrayList<Integer> a1 = new ArrayList<>();
			while (l < r) {

				// st.insert(0, a[r] + " ");
				// st.insert(0, a[l] + " ");
				a1.add(a[r]);
				a1.add(a[l]);
				l++;
				r--;
			}
			if (n % 2 != 0) {
				a1.add(a[r]);
			}
			for(int i=a1.size()-1;i>=0;i--) {
				sb.append(a1.get(i)+" ");
			}
			sb.append("\n");
		}
		System.out.println(sb);
	}

	private static class pair implements Comparable<pair> {
		int x;
		int y;

		pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		public int compareTo(pair o) {
			if (this.y == o.y) {
				return this.x - o.x;
			}
			return this.y - o.y;
		}
	}
}