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:
"""
    Author : thekushalghosh
    Team   : CodeDiggers
"""
import sys,math,cmath,time
start_time = time.time()
################# ---- USER DEFINED INPUT FUNCTIONS ---- #################
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(s[:len(s) - 1])
def invr():
    return(map(int,input().split()))
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################

def solve():
    n = inp()
    a = inlt()
    q = []
    a.sort()
    for i in range(math.ceil(len(a) / 2)):
        q.append(a[n - i - 1])
        q.append(a[i])
    if len(a) % 2 == 1:
        q.pop()
    q.reverse()
    print(*q)
    
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
ONLINE_JUDGE = __debug__
if not ONLINE_JUDGE:
    sys.stdin = open('input.txt','r')
    sys.stdout = open('output.txt','w')
else:
    input = sys.stdin.readline
t = 1
t = inp()
for tt in range(t):
    solve()
if not ONLINE_JUDGE:
    print("Time Elapsed:",time.time() - start_time,"seconds")
sys.stdout.close()