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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class test {
    static void removeDups(int[] arr, int n)  
    { 
  
        // Hash map which will store the 
        // elements which has appeared previously. 
        HashMap<Integer,  
                Boolean> mp = new HashMap<>(); 
  
        for (int i = 0; i < n; ++i) 
        { 
  
            // Print the element if it is not 
            // there in the hash map 
            if (mp.get(arr[i]) == null) 
                System.out.print(arr[i] + " "); 
  
            // Insert the element in the hash map 
            mp.put(arr[i], true); 
        } 
    }
    

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
    Scanner in=new Scanner(System.in);
        int i,j,ni,n,l,x,y,z,c=0;
        boolean flag=true;
        int tc=in.nextInt();
        for(i=0;i<tc;i++)
        {
            n=in.nextInt();
           int a[]=new int[2*n];
           
           c=0;
           for(j=0;j<2*n;j++)
           {a[j]=in.nextInt();
            }
           removeDups(a,2*n);
           System.out.println();
          
        
    
    }}
}