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 math, sys
from collections import defaultdict, Counter, deque

INF = float('inf')
MOD = 1000000007

def gcd(a, b):
	while b:
		a, b = b, a%b
	return a

def isPrime(n):
	if (n <= 1): 
		return False
	i = 2
	while i ** 2 <= n:
		if n % i == 0:
			return False
		i += 1
	return True

def primeFactors(n):
	factors = []
	i = 2
	while i ** 2 <= n:
		while n % i == 0:
			factors.append(i)
			n //= i 
		i += 1
	if n > 1:
		factors.append(n)
	return factors

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

def array():
	return list(map(int, input().split()))

ans = 0

def dfs(visited, node, digits, tree):
	if not visited[node]:
		visited[node] = True
		digits += str(node)
		leafNode = True
		
		for child in tree[node]:
			leafNode = False
			dfs(visited, child, digits, tree)
			
		if leafNode:
			global ans
			ans += int(digits)


def main():
	n = int(input())
	arr = array()
	m = defaultdict(int)
	ans = []
	for i in range(n * 2):
		if m[arr[i]] == 0:
			ans.append(arr[i])
			m[arr[i]] += 1 
	print(*ans)

if __name__ == "__main__":
	t = int(input())
	# t = 1
	for _ in range(t):
		main()