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:
# DEFINING SOME GOOD STUFF
from math import *
import threading
import sys
from collections import *
import itertools
sys.setrecursionlimit(300000)
# threading.stack_size(10**5)  # remember it cause mle


mod = 10 ** 9
inf = 10 ** 15
yes = 'YES'
no = 'NO'

# ------------------------------FASTIO----------------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")



# _______________________________________________________________#

def npr(n, r):
    return factorial(n) // factorial(n - r) if n >= r else 0


def ncr(n, r):
    return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0


def lower_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1

    while (start <= end):
        middle = (end + start) // 2
        if li[middle] >= num:
            answer = middle
            end = middle - 1
        else:
            start = middle + 1
    return answer  # min index where x is not less than num


def upper_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1

    while (start <= end):
        middle = (end + start) // 2

        if li[middle] <= num:
            answer = middle
            start = middle + 1

        else:
            end = middle - 1
    return answer  # max index where x is not greater than num


def abs(x):
    return x if x >= 0 else -x


def binary_search(li, val):
    # print(lb, ub, li)
    ans = -1
    lb = 0
    ub = len(li) - 1
    while (lb <= ub):
        mid = (lb + ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid - 1
        elif val > li[mid]:
            lb = mid + 1
        else:
            ans = mid  # return index
            break
    return ans


def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far


def pref(li):
    pref_sum = [0]
    for i in li:
        pref_sum.append(pref_sum[-1] + i)
    return pref_sum


def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    li = []
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1

    for p in range(2, len(prime)):
        if prime[p]:
            li.append(p)
    return li


def primefactors(n):
    factors = []
    while (n % 2 == 0):
        factors.append(2)
        n //= 2
    for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
        while n % i == 0:
            factors.append(i)
            n //= i
    if n > 2:  # incase of prime
        factors.append(n)
    return factors


def prod(li):
    ans = 1
    for i in li:
        ans *= i
    return ans


def dist(a, b):
    d = abs(a[1] - b[1]) + abs(a[2] - b[2])
    return d


def power_of_n(x, n):
    cnt = 0
    while (x % n == 0):
        cnt += 1
        x //= n
    return cnt


def ask(l, r):
    if l == r:
        return -1
    print("?", l, r)
    sys.stdout.flush()
    return int(input())


# _______________________________________________________________#


# def main():

# for _ in range(1):
for _ in range(int(input())):
    n = int(input())
    # n, b = map(int, input().split())
    a = list(map(int, input().split()))
    # b = list(map(int, input().split()))
    # c = list(map(int, input().split()))
    # s = list(input())
    # s = input()
    # t = input()
    # c = list(map(int,input().split()))
    # adj = graph(n,m)
    a.sort()
    ans = []
    l, r = 0, n-1
    while l < r:
        ans.append(a[r])
        ans.append(a[l])
        l += 1
        r -= 1
    if l == r:
        ans.append(a[r])
    ans.reverse()
    print(*ans)

























'''		
t = threading.Thread(target=main)
t.start()
t.join()
'''