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:
def solve_odd_sorted(a):
    n = len(a)
    assert n % 2 == 1
    median_pos = n // 2
    up_first = abs(a[median_pos + 1] - a[median_pos]) < abs(
        a[median_pos - 1] - a[median_pos]
    )
    res = [a[median_pos]]
    if not up_first:
        for i in range(1, n // 2 + 1):
            res.append(a[median_pos - i])
            res.append(a[median_pos + i])
    else:
        for i in range(1, n // 2 + 1):
            res.append(a[median_pos + i])
            res.append(a[median_pos - i])
    return res


def solve(a):
    n = len(a)
    a = sorted(a)

    if n % 2 == 1:
        median = a[n // 2]
    else:
        median = (a[n // 2 - 1] + a[n // 2]) / 2

    if len(a) % 2 == 1:
        return solve_odd_sorted(a)
    return solve_odd_sorted(a[: n // 2] + [median] + a[n // 2 :])[1:]


def main():
    T = int(input())
    for _ in range(T):
        n = int(input())
        a = list(map(int, input().split()))
        print(" ".join(map(str, solve(a))))


if __name__ == "__main__":
    main()