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 java.util.*;
import java.io.*;

public class Yash 
{
	static void TopK(ArrayList<Integer> arr, int k) 
	{
		LinkedHashSet<Integer> a = new LinkedHashSet<>();
		int n = arr.size();
		TreeMap<Integer, Integer> hp = new TreeMap<>(Collections.reverseOrder());
		for(int i=0; i<n; i++)
		{
			if(hp.containsKey(arr.get(i)))
				hp.put(arr.get(i), hp.get(arr.get(i))+1);
			else
				hp.put(arr.get(i), 1);
		}
		TreeMap<Integer, Integer> tr = new TreeMap<>();
		PriorityQueue<Integer> hps = new PriorityQueue<>(Collections.reverseOrder());
		for (Map.Entry entry : hp.entrySet())
		{
		    hps.add((int)entry.getValue());
		}
		int count =0;
		boolean doublebreak = true;
		while(!hps.isEmpty() && doublebreak)
		{
			count++;
			if(count>k)
				break;
			for(int key: hp.keySet()) 
			{
			    if(hp.get(key).equals(hps.peek())) 
			    {
			    	if(a.size()==k)
			    	{
			    		doublebreak = false;
			    		break;
			    	}
			        a.add(key);
			    }
			    
			}
			hps.remove();
		}
		ArrayList<Integer> fin = new ArrayList<>();
		Iterator<Integer> i = a.iterator();
		while(i.hasNext())
			fin.add(i.next());
		System.out.println(hp);
		System.out.println(hps);
		System.out.println(a);
		System.out.println(fin);
	}
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0)
		{
			int n = sc.nextInt();
			LinkedHashSet<Integer> hs = new LinkedHashSet<>();
			int a[] = new int[2*n];
			for(int i=0; i<2*n; i++)
			{
				a[i] = sc.nextInt();
				hs.add(a[i]);
			}
			Iterator<Integer> i = hs.iterator();
			while(i.hasNext())
				System.out.print(i.next() + " ");
			System.out.println();
		}
	}
}