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:
"""
  ____          _      _____
 / ___|___   __| | ___|  ___|__  _ __ ___ ___  ___
| |   / _ \ / _` |/ _ \ |_ / _ \| '__/ __/ _ \/ __|
| |__| (_) | (_| |  __/  _| (_) | | | (_|  __/\__ \
 \____\___/ \__,_|\___|_|  \___/|_|  \___\___||___/

"""
"""
░░██▄░░░░░░░░░░░▄██
░▄▀░█▄░░░░░░░░▄█░░█░
░█░▄░█▄░░░░░░▄█░▄░█░
░█░██████████████▄█░
░█████▀▀████▀▀█████░
▄█▀█▀░░░████░░░▀▀███
██░░▀████▀▀████▀░░██
██░░░░█▀░░░░▀█░░░░██
███▄░░░░░░░░░░░░▄███
░▀███▄░░████░░▄███▀░
░░░▀██▄░▀██▀░▄██▀░░░
░░░░░░▀██████▀░░░░░░
░░░░░░░░░░░░░░░░░░░░
"""

import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt, pi, cos, sin, ceil, log2
from bisect import bisect_right, bisect_left

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)


def factorial(n):
    if n == 0:
        return 1
    return n * factorial(n - 1)


def ncr(n, r):
    r = min(r, n - r)
    numer = reduce(op.mul, range(n, n - r, -1), 1)
    denom = reduce(op.mul, range(1, r + 1), 1)
    return numer // denom  # or / in Python 2


def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return len(set(factors))


def isPowerOfTwo(x):
    return (x and (not(x & (x - 1))))


def factors(n):
    return list(set(reduce(list.__add__, ([i, n // i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))


def DFS(root, g, vis, ans):
    vis[root] = 1
    # print(g[root])
    for c in g[root]:
        if vis[c] == 0:
            ans.append(c)
            DFS(c, g, vis, ans)

MOD = 1000000007
PMOD = 998244353
T = 1
T = int(stdin.readline())
for _ in range(T):
    # x, y, z = list(map(int, stdin.readline().rstrip().split()))
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().rstrip().split()))
    # q = int(stdin.readline())
    # c = list(map(int, stdin.readline().rstrip().split()))
    # s = str(stdin.readline().strip('\n'))
    d = {}
    x = []
    for i in range(2 * n):
        if a[i] not in d:
            x.append(a[i])
            d[a[i]] = 1
        else:
            d[a[i]] += 1
    print(*x)