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 math
from decimal import Decimal
import heapq
import copy
import heapq
from collections import deque
def na():
	n = int(input())
	b = [int(x) for x in input().split()]
	return n,b
 
	
def nab():
	n = int(input())
	b = [int(x) for x in input().split()]
	c = [int(x) for x in input().split()]
	return n,b,c
 
		
def dv():
	n, m = map(int, input().split())
	return n,m
 
 
def da():
	n, m = map(int, input().split())
	a = list(map(int, input().split()))
	return n,m, a 
 
 
def dva():
	
	n, m = map(int, input().split())
	a = [int(x) for x in input().split()]
	b = [int(x) for x in input().split()]
	return n,m,b
 
 
def eratosthenes(n): 
	sieve = list(range(n + 1))
	for i in sieve:
		if i > 1:
			for j in range(i + i, len(sieve), i):
				sieve[j] = 0
	return sorted(set(sieve))
 
 
def lol(lst,k):
	k=k%len(lst)
	ret=[0]*len(lst)
	for i in range(len(lst)):
		if i+k<len(lst) and i+k>=0:
			ret[i]=lst[i+k]
		if i+k>=len(lst):
			ret[i]=lst[i+k-len(lst)]
		if i+k<0:
			ret[i]=lst[i+k+len(lst)]
	return(ret)
def nm():
	n = int(input())
	b = [int(x) for x in input().split()]
	m = int(input())
	c = [int(x) for x in input().split()]
	return n,b,m,c
 
 
def dvs():
	n = int(input())
	m = int(input())
	return n, m 
 
 
def fact(n):
	tc = []
	ans = {}
	d = 2
	while d * d <= n:
		if n % d == 0:
			tc.append(d)
			n //= d
		else:
			d += 1
	if n > 1:
		tc.append(n)
	for i in tc:
		ans[i] = ans.get(i, 0) + 1
	return ans
 
 
for _ in range(int(input())):
		n, a = na()
		d = {}
		for i in range(1, n + 1):
			d[i] = 0
		ans = []
		for i in a:
			if d[i] == 0:
				d[i] += 1
			else:
				ans.append(i)
		print(*ans)
		
