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 
/* package whatever; // don't place package name! */
 
import java.util.*;
import java.math.BigInteger; 
import java.lang.*;
import java.io.*;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.lang.Math; 
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
 
/* Name of the class has to be "Main" only if the class is public. */
public class bbg
{
	public static int result;
 
public static ArrayList<Integer> [] graph;
 
public static int[]cats;
 public static String[]vector;
 
 
public static int vizitat[];
public static int x;
//public static HashMap<String, Integer> map2;
 
 
	public static void main (String[] args) throws IOException {
 
Scanner input = new Scanner(System.in);
 
HashMap<Integer, Integer> contor1= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> contor2= new HashMap<Integer, Integer>();


HashMap<Integer, Integer> map= new HashMap<Integer, Integer>();
HashMap<String, Integer> litere= new HashMap<String, Integer>();
HashMap<String, Integer> combinari= new HashMap<String, Integer>();


 litere.put("a",1);
 litere.put("e",1);
litere.put("i",1);
litere.put("o",1);

litere.put("u",1);
 BigInteger numar_initial; 
 BigInteger primul; 
 BigInteger doilea;
  BigInteger produs;
  BigInteger unitatea;
  
  int teste=input.nextInt();
 int min,max;
 
  for (int t=1; t<=teste;t++){
  
  
  int n=input.nextInt();
  int vector[]=new int[n+1];
  int contor=1;
  for(int i=0;i<2*n;i++)
  {
   int numar=input.nextInt();
   
   if (vector[numar]==0){ if (contor==n) System.out.println(numar + " ");
    
    else System.out.print(numar + " ");
    
    contor++;
    
 
    
    vector[numar]=1;
    
   }
   
   
   
   
   else vector[numar]++;
   
  }

 
 
 }
 
 
}

}