Problem: A permutation of length n is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] are permutations, and [1, 1], [0, 1], [2, 2, 1, 4] are not.

There was a permutation p[1 ... n]. It was merged with itself. In other words, let's take two instances of p and insert elements of the second p into the first maintaining relative order of elements. The result is a sequence of the length 2n.

For example, if p=[3, 1, 2] some possible results are: [3, 1, 2, 3, 1, 2], [3, 3, 1, 1, 2, 2], [3, 1, 3, 1, 2, 2]. The following sequences are not possible results of a merging: [1, 3, 2, 1, 2, 3], [3, 1, 2, 3, 2, 1], [3, 3, 1, 2, 2, 1].

For example, if p=[2, 1] the possible results are: [2, 2, 1, 1], [2, 1, 2, 1]. The following sequences are not possible results of a merging: [1, 1, 2, 2], [2, 1, 1, 2], [1, 2, 2, 1].

Your task is to restore the permutation p by the given resulting sequence a. It is guaranteed that the answer exists and is unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 400) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 50) — the length of permutation. The second line of the test case contains 2n integers a_1, a_2, ..., a_{2n} (1 ≤ a_i ≤ n), where a_i is the i-th element of a. It is guaranteed that the array a represents the result of merging of some permutation p with the same permutation p.

Output

For each test case, print the answer: n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n), representing the initial permutation. It is guaranteed that the answer exists and is unique.

Example

Input


5
2
1 1 2 2
4
1 3 1 4 3 4 2 2
5
1 2 1 2 3 4 3 5 4 5
3
1 2 3 1 2 3
4
2 3 2 4 1 3 4 1


Output


1 2 
1 3 4 2 
1 2 3 4 5 
1 2 3 
2 3 4 1 

Solution (in PYTHON3):

Solution:
#
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#         | |   / _` |/ _` |/ _ \ '_ \|_  / _` |
#         | |__| (_| | (_| |  __/ | | |/ / (_| |
#          \____\____|\____|\___|_| |_/___\____|
#
#      GNU Affero General Public License v3.0
#    ------------------------------------------------
#    Author   : prophet
#    Created  : 2020-07-17 10:46:23.322074
#    UUID     : BLkyVuW845Lb834H
#    ------------------------------------------------
#
production = True

import sys, math, collections

def input(input_format = 0, multi = 0):

    if multi > 0: return [input(input_format) for i in range(multi)]
    else:
        next_line = sys.stdin.readline()[:-1]

        if input_format >= 10:
            use_list = False
            input_format = int(str(input_format)[-1])
        else: use_list = True

        if input_format == 0: formatted_input = [next_line]
        elif input_format == 1: formatted_input = list(map(int, next_line.split()))
        elif input_format == 2: formatted_input = list(map(float, next_line.split()))
        elif input_format == 3: formatted_input = list(next_line)
        elif input_format == 4: formatted_input = list(map(int, list(next_line)))
        elif input_format == 5: formatted_input = next_line.split()
        else: formatted_input = [next_line]

        return formatted_input if use_list else formatted_input[0]

def out(output_line, output_format = 0, newline = True):

    formatted_output = ""

    if output_format == 0: formatted_output = str(output_line)
    elif output_format == 1: formatted_output = " ".join(map(str, output_line))
    elif output_format == 2: formatted_output = "\n".join(map(str, output_line))

    print(formatted_output, end = "\n" if newline else "")

def log(*args):
    if not production:
        print("$$$", end = "")
        print(*args)

enu = enumerate

ter = lambda a, b, c: b if a else c

ceil = lambda a, b: -(-a // b)

def mapl(iterable, format = 0):
    
    if format == 0: return list(map(int, iterable))
    elif format == 1: return list(map(str, iterable))
    elif format == 2: return list(map(list, iterable))
#
#   >>>>>>>>>>>>>>> START OF SOLUTION <<<<<<<<<<<<<<
#


def solve():

    n = input(11)
    a = input(1)

    b = set()
    y = []

    for i in a:
        if i not in b:
            y.append(i)
            b.add(i)

    out(y, 1)

    return


for i in range(input(11)): solve()
# solve()

#
#   >>>>>>>>>>>>>>>> END OF SOLUTION <<<<<<<<<<<<<<<
#