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 PYTHON):

Solution:
"""Template for Python Competitive Programmers prepared by pa.n.ik, kabeer seth and Mayank Chaudhary """

# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""

def prepare_factorial():
    fact = [1]
    for i in range(1, 100005):
        fact.append((fact[-1] * i) % mod)
    ifact = [0] * 100005
    ifact[100004] = pow(fact[100004], mod - 2, mod)
    for i in range(100004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod

    return fact, ifact

"""uncomment next 4 lines while doing recursion based question"""
# import threading
# threading.stack_size(1<<27)
import sys
# sys.setrecursionlimit(10000)


"""uncomment modules according to your need"""

from bisect import bisect_left, bisect_right, insort
from math import floor, ceil, sqrt, degrees, atan, pi, log, sin, radians
from heapq import heappop, heapify, heappush
from collections import Counter, defaultdict, deque


def modinv(n, p):
    return pow(n, p - 2, p)

def ncr(n, r, fact, ifact):    # for using this uncomment the lines calculating fact and ifact
    t = (fact[n] * (ifact[r] * ifact[n-r]) % mod) % mod
    return t


def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

"""*****************************************************************************************"""

def GCD(x, y):
    while y:
        x, y = y, x % y
    return x

def lcm(x, y):
    return (x * y)//(GCD(x, y))

def get_xor(n):
    return [n, 1, n+1, 0][n % 4]

def fast_expo(a, b):
    res = 1
    while b:
        if b&1:
            res = (res * a)
            res %= MOD
            b -= 1
        else:
            a = (a* a)
            a %= MOD
            b>>=1
    res %= MOD
    return res

def get_n(P):    # this function returns the maximum n for which Summation(n) <= Sum

    ans = (-1 + sqrt(1 + 8*P))//2
    return ans

""" ********************************************************************************************* """

def main():

    T = int(input())
    while T:
        n = int(input())
        Arr = get_array()

        ans = []
        myset = set()

        for i in Arr[::-1]:
            if i in myset:
                continue
            ans.append(i)
            myset.add(i)

        ans.reverse()
        print(*ans)
        T -= 1




""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

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

""" main function"""

if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()