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

/**
  * Built using my Brain
  * Actual solution is at the bottom
  *
  * @author Lenard Hoffstader
*/

public class cfjava
{
  public static void main(String[] args)
  {
    OutputStream outputStream = System.out;
    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(outputStream);
    PROBLEM solver = new PROBLEM();
    int t=1;
    t = in.nextInt();
    for(int i=0;i<t;i++){
      solver.solve(in, out);
    }
    out.close();
  }
  static class PROBLEM
  {
    public void solve(FastReader in,PrintWriter out)
    {
      int n = in.nextInt(),x=0;
      HashMap<Integer,Boolean> mp = new HashMap<>();
      ArrayList<Integer>  v1 = new ArrayList<>();
      for(int i=0;i<2*n;i++){
        x = in.nextInt();
        if(!mp.containsKey(x)){
          v1.add(x);
          mp.put(x,true);
        }
      }
      for(int xx:v1)out.print(xx+" ");
      out.println();

    }
  }
  
  static class FastReader 
  {
    BufferedReader br; 
    StringTokenizer st;

    public FastReader() 
    { 
      br = new BufferedReader(new InputStreamReader(System.in)); 
    } 

    String next() 
    { 
      while (st == null || !st.hasMoreElements()) 
      {
        try
        { 
          st = new StringTokenizer(br.readLine()); 
        } 
        catch (IOException  e) 
        { 
          e.printStackTrace(); 
        } 
      } 
      return st.nextToken(); 
    } 

    int nextInt(){return Integer.parseInt(next());}

    long nextLong(){return Long.parseLong(next());} 

    double nextDouble(){return Double.parseDouble(next());}

    boolean nextBoolean(){return !(nextInt()==0);}
    // boolean nextBoolean(){return Boolean.parseBoolean(next());}

    String nextLine() 
    { 
      String str = ""; 
      try
      { 
        str = br.readLine(); 
      } 
      catch (IOException e){e.printStackTrace();} 
      return str; 
    } 
  }
}