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, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
#from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
from fractions import Fraction
#sys.setrecursionlimit(100000)
#INF = float('inf')
mod = int(1e9)+7


for t in range(int(data())):
    n=int(data())
    a=mdata()
    d=dd(int)
    ans=[]
    for i in range(2*n):
        d[a[i]]+=1
        if d[a[i]]==2:
            ans.append(a[i])
            d[a[i]]=0
    outl(ans)