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 
'''     J A I ~ S H R E E ~ R A M     '''

# Title: cf-1385-b.py
# created on: 17-07-2020 at 20:12:42
# Creator & Template : Udit Gupta "@luctivud"
# https://github.com/luctivud
# https://www.linkedin.com/in/udit-gupta-1b7863135/


import math; from collections import *
import sys; from functools import reduce
from itertools import groupby

# sys.setrecursionlimit(10**6)

def get_ints(): return map(int, input().strip().split())
def get_list(): return list(get_ints())
def get_string(): return list(input().strip().split())
def printxsp(*args): return print(*args, end="")
def printsp(*args): return print(*args, end=" ")


DIRECTIONS = [[0, 1], [0, -1], [1, 0], [1, -1]] #up, down, right, left
NEIGHBOURS = [(i, j) for i in range(-1, 2) for j in range(-1, 2) if (i!=0 or j!=0)]


OrdUnicode_a = ord('a'); OrdUnicode_A = ord('A')
CAPS_ALPHABETS = {chr(i+OrdUnicode_A) : i for i in range(26)}
SMOL_ALPHABETS = {chr(i+OrdUnicode_a) : i for i in range(26)}


MOD_JOHAN = int(1e9)+7; MOD_LIGHT = 998244353; INFINITY = float('inf')
MAXN_EYEPATCH = int(1e5)+1; MAXN_FULLMETAL = 501

# Custom input output is now piped through terminal commands.


for _testcases_ in range(int(input())): 
    n = int(input())
    li = get_list()
    found = [False] * (n+1)
    for i in li:
        if not found[i]:
            found[i] = True
            printsp(i)
    print()





'''
THE LOGIC AND APPROACH IS MINE ( UDIT GUPTA )
Link may be copy-pasted here, otherwise.
'''
