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.ArrayList;
import java.awt.Dimension;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Objects;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.SortedMap;
import java.util.SortedSet;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.regex.Pattern;
import java.util.stream.Collectors;

import com.sun.org.apache.xpath.internal.operations.Mod;

import java.util.Comparator;
import java.util.Date;
import java.util.HashMap;
import java.util.HashSet;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.text.SimpleDateFormat;
import java.time.LocalDate;
public class myFile
{	
	public void finalize()
	{
		System.out.println("Cl");
	}
	public static void main(String args[]) 
	{ 
	  int test,n,i;
	  LinkedHashSet<Integer> list = new LinkedHashSet<Integer>();
	  Scanner sc = new Scanner(System.in);
	  test = sc.nextInt();
	  while(test>0)
	  {
		  test--;
		  list.clear();
		  n = sc.nextInt();
		  for(i=0;i<(2*n);i++)
			  list.add(sc.nextInt());
		  
		  Iterator<Integer> iter = list.iterator();
		  int count=0;
		  while(iter.hasNext())
		  {
			  if(count==0)
			  System.out.print(iter.next());
			  else
				  System.out.print(" "+iter.next());
			  
			  count++;
		  }
		  System.out.println();
	  }
		
	}
}
