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 PYTHON3):

Solution:
"""
steps:
-> sort array
-> find median
-> median (i) goes first than i + 1, i - 1, i + 2, i - 2 etc.
"""
def quicker_partition(tab, p, r):
    x = tab[r]
    i = p - 1
    k = r - 1
    j = p
    while j <= k:
        if tab[j] < x:
            i += 1
            tab[i], tab[j] = tab[j], tab[i]
            j += 1
        elif tab[j] == x:
            tab[k], tab[j] = tab[j], tab[k]
            k -= 1
        else:
            j += 1
    for l in range(r - k):
        tab[l + i + 1], tab[k + 1 + l] = tab[k + 1 + l], tab[l + i + 1]
    return i, i + r - k + 1


def quicker_sort(tab, p, r):
    while p < r:
        left, right = quicker_partition(tab, p, r)
        if left - p < r - right:
            quicker_sort(tab, p, left)
            p = right
        else:
            quicker_sort(tab, right, r)
            r = left



def sort_out_array(tab):
    quicker_sort(tab, 0, len(tab) - 1)
    i = (len(tab) - 1) // 2
    print(tab[i], end=" ")
    j = i - 1
    k = i + 1
    while j > -1 or k < len(tab):
        if k < len(tab):
            print(tab[k], end=" ")
            k += 1
        if j > -1:
            print(tab[j], end=" ")
            j -= 1
    print()


def get_input():
    number_of_problems = int(input())
    arrays = [0] * number_of_problems
    for i in range(number_of_problems):
        length_of_array = int(input())
        temp = input().split(sep=" ")
        for j in range(length_of_array):
            temp[j] = int(temp[j])
        arrays[i] = temp
    return arrays


if __name__ == "__main__":
    arrays_to_sort = get_input()
    for array in arrays_to_sort:
        sort_out_array(array)