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 
from sys import stdin, stdout
from sys import maxsize
input = stdin.readline
# def print(n):
#    stdout.write(str(n)+'\n')


def solve():
    pass


test = 1
test = int(input().strip())
for t in range(0, test):
    n = int(input().strip())
    # s = list(input().strip())                                                        # String Input, converted to mutable list.
    # n, k = list(map(int, input().strip().split()))
    a = [int(x) for x in input().strip().split()]
    ans = []
    visited = set()
    for el in a:
        if(el not in visited):
            ans += [el]
            visited.add(el)

    def onedarr(a):
        for el in a:
            print(el, end=' ')
        print()
    onedarr(ans)
    # brr = [list(map(int,input().strip().split())) for i in range(rows)]              # 2D array row-wise input
    ans = solve()


'''
rows, cols = (5, 5)
arr = [[0]*cols for j in range(rows)]                                         # 2D array initialization
rev_str=s[::-1]                                                               # To reverse given string
b=input().strip().split()                                                     # list created by spliting about spaces
brr = [[int(b[cols*i+j]) for j in range(cols)] for i in range(rows)]          # 2D array Linear Input
rows,cols=len(brr),len(brr[0])                                                # no of rows/cols for 2D array
arr.sort(key = lambda x : x[1])                                               # sort list of tuples by 2nd element, Default priority - 1st Element then 2nd Element
s=set()                                                                       # empty set
a=maxsize                                                                     # initializing infinity
b=-maxsize                                                                    # initializing -infinity
mapped=list(map(function,input))                                              # to apply function to list element-wise
try:                                                                          # Error handling
    #code 1
except:                                                                       # ex. to stop at EOF
    #code 2 , if error occurs
'''
