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:

from bisect import bisect_left
from math import ceil
from math import radians
from heapq import heapify, heappush, heappop
import bisect
from math import pi
from collections import deque
from math import factorial
from math import log, ceil
from collections import defaultdict
from math import *
from sys import stdin, stdout
import itertools
import os
import sys
import threading
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import *
# from math import ceil, floor, log, sqrt, factorial, pow, pi, gcd
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
from fractions import Fraction
mod = int(pow(10, 9)+7)
# mod = 998244353


def ii(): return int(input())


def si(): return str(input())


def mi(): return map(int, input().split())


def li1(): return list(mi())


def fii(): return int(stdin.readline())


def fsi(): return str(stdin.readline())


def fmi(): return map(int, stdin.readline().split())


def fli(): return list(fmi())


abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
       'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
       'z': 25}


def getKey(item): return item[0]


def sort2(l): return sorted(l, key=getKey)


def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]


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


def decimalToBinary(n): return bin(n).replace("0b", "")


def ntl(n): return [int(i) for i in str(n)]


def powerMod(x, y, p):
    res = 1
    x %= p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res


graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000


def bfs(d, v):
    q = []
    q.append(v)
    visited[v] = 1
    while len(q) != 0:
        x = q[0]
        q.pop(0)
        for i in d[x]:
            if visited[i] != 1:
                visited[i] = 1
                q.append(i)
        print(x)


def make_graph(e):
    d = {}
    for i in range(e):
        x, y = mi()
        if x not in d:
            d[x] = [y]
        else:
            d[x].append(y)
        if y not in d:
            d[y] = [x]
        else:
            d[y].append(x)
    return d


def gr2(n):
    d = defaultdict(list)
    for i in range(n):
        x, y = mi()
        d[x].append(y)
    return d


def connected_components(graph):
    seen = set()

    def dfs(v):
        vs = set([v])
        component = []
        while vs:
            v = vs.pop()
            seen.add(v)
            vs |= set(graph[v]) - seen
            component.append(v)
        return component

    ans = []
    for v in graph:
        if v not in seen:
            d = dfs(v)
            ans.append(d)
    return ans


def primeFactors(n):
    s = set()
    while n % 2 == 0:
        # s.add(2)
        n = n // 2
    for i in range(3, int(sqrt(n)) + 1, 2):
        while n % i == 0 and i % 2 == 1:
            s.add(i)
            n = n // i
    if n > 2 and n % 2 == 1:
        s.add(n)
    return s


def find_all(a_str, sub):
    start = 0
    while True:
        start = a_str.find(sub, start)
        if start == -1:
            return
        yield start
        start += len(sub)


def SieveOfEratosthenes(n, isPrime):
    isPrime[0] = isPrime[1] = False
    for i in range(2, n):
        isPrime[i] = True
    p = 2
    while (p * p <= n):
        if (isPrime[p] == True):
            i = p * p
            while (i <= n):
                isPrime[i] = False
                i += p
        p += 1
    return isPrime


def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l, r, c in edges:
        g[l].append((c, r))

    q, seen, mins = [(0, f, ())], set(), {f: 0}
    while q:
        (cost, v1, path) = heappop(q)
        if v1 not in seen:
            seen.add(v1)
            path = (v1, path)
            if v1 == t:
                return (cost, path)

            for c, v2 in g.get(v1, ()):
                if v2 in seen:
                    continue
                prev = mins.get(v2, None)
                next = cost + c
                if prev is None or next < prev:
                    mins[v2] = next
                    heappush(q, (next, v2, path))
    return float("inf")


def binsearch(a, l, r, x):
    while l <= r:
        mid = l + (r-1)//2
        if a[mid]:
            return mid
        elif a[mid] > x:
            l = mid-1
        else:
            r = mid+1
    return -1


# def input():
#    return stdin.buffer.readline()


def readTree(n):
    adj = [set() for _ in range(n)]
    for _ in range(n-1):
        u, v = map(int, input().split())
        adj[u-1].add(v-1)
        adj[v-1].add(u-1)
    return adj


def treeOrderByDepth(n, adj, root=0):
    parent = [-2] + [-1]*(n-1)
    ordered = []
    q = deque()
    q.append(root)
    depth = [0] * n
    while q:
        c = q.popleft()
        ordered.append(c)
        for a in adj[c]:
            if parent[a] == -1:
                parent[a] = c
                depth[a] = depth[c] + 1
                q.append(a)
    return (ordered, parent, depth)


def isSubSequence(str1, str2):
    m = len(str1)
    n = len(str2)

    j = 0    # Index of str1
    i = 0    # Index of str2

    # Traverse both str1 and str2
    # Compare current character of str2 with
    # first unmatched character of str1
    # If matched, then move ahead in str1

    while j < m and i < n:
        if str1[j] == str2[i]:
            j = j+1
        i = i + 1

    # If all characters of str1 matched, then j is equal to m
    return j == m


def nextPowerOf2(n):
    count = 0

    # First n in the below
    # condition is for the
    # case where n is 0
    if (n and not(n & (n - 1))):
        return n

    while(n != 0):
        n >>= 1
        count += 1

    return 1 << count


def cou(n):
    c = 0
    while n > 1:
        c += 1
        n //= 2
    return c


def sortsec(l):
    return sorted(l, key=lambda x: x[1])


def BinarySearch(a, x):
    i = bisect_left(a, x)
    if i:
        return (i-1)
    else:
        return - 1


for _ in range(ii()):
    n = ii()
    l = li1()
    out = []
    b = [0] * (n + 1)
    for i in l:
        if b[i] == 0:
            out.append(i)
            b[i] = 1
    print(*out)