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 
import bisect
import decimal
from decimal import Decimal
import os
from collections import Counter
import bisect
from collections import defaultdict
import math
import random
import heapq 
from math import sqrt
import sys
from functools import reduce, cmp_to_key
from collections import deque
import threading
from itertools import combinations
from io import BytesIO, IOBase
from itertools import accumulate

# sys.setrecursionlimit(200000)
mod = 10**9+7
# mod = 998244353
decimal.getcontext().prec = 46

def primeFactors(n): 
    prime = set()
    while n % 2 == 0: 
        prime.add(2)
        n = n//2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            prime.add(i)
            n = n//i 
    if n > 2: 
        prime.add(n)
    return list(prime)
    
def getFactors(n) : 
    factors = []
    i = 1
    while i <= math.sqrt(n): 
        if (n % i == 0) : 
            if (n // i == i) : 
                factors.append(i)
            else : 
                factors.append(i)
                factors.append(n//i)
        i = i + 1
    return factors
    
def modefiedSieve(): 
    mx=10**7+1
    sieve=[-1]*mx
    for i in range(2,mx):
        if sieve[i]==-1:
            sieve[i]=i
            for j in range(i*i,mx,i):
                if sieve[j]==-1:
                    sieve[j]=i
    return sieve
    
def SieveOfEratosthenes(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    num = []
    for p in range(2, n+1): 
        if prime[p]: 
            num.append(p)
    return num

def lcm(a,b):
    return (a*b)//math.gcd(a,b)

def sort_dict(key_value):
    return sorted(key_value.items(), key = lambda kv:(kv[1], kv[0]), reverse=True)

def list_input():
    return list(map(int,input().split()))
    
def num_input():
    return map(int,input().split())
    
def string_list():
    return list(input())
    
def decimalToBinary(n):  
    return bin(n).replace("0b", "")  
    
def binaryToDecimal(n): 
    return int(n,2) 
    
def DFS(n,s,adj):              
    visited = [False for i in range(n+1)]  
    stack = [] 
    stack.append(s)  
  
    while (len(stack)):  
        s = stack[-1]  
        stack.pop() 
  
        if (not visited[s]):  
            visited[s] = True 
            
        for node in adj[s]:  
            if (not visited[node]):  
                stack.append(node)  
        
def maxSubArraySum(a,size): 
    max_so_far = -sys.maxsize - 1
    max_ending_here = 0
    start = 0
    end = 0
    s = 0
  
    for i in range(0,size): 
        max_ending_here += a[i] 
        if max_so_far < max_ending_here: 
            max_so_far = max_ending_here 
            start = s 
            end = i 
  
        if max_ending_here < 0: 
            max_ending_here = 0
            s = i+1
    
    return max_so_far,start,end
    
def lis(arr): 
    n = len(arr) 
    lis = [1]*n 
    for i in range (1 , n): 
        for j in range(0 , i): 
            if arr[i] >= arr[j] and lis[i]< lis[j] + 1 : 
                lis[i] = lis[j]+1
    maximum = 0
    for i in range(n): 
        maximum = max(maximum , lis[i]) 
  
    return maximum 
        
def solve():
    n = int(input())
    arr = list_input()
    ans = []
    for i in range(2*n):
        if not(arr[i] in ans):
            ans.append(arr[i])
    print(*ans)
    
t = 1
t = int(input())
for _ in range(t):
    solve()


    
    
    
    
