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 
// template halnix16


import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

import static java.lang.Math.*;

public class Main {


    static BufferedReader br;

    public static void main(String[] args) throws Exception {
        // write your code here
        br = new BufferedReader(new InputStreamReader(System.in));

        int t = (int)in();
        while (t-->0){

            int n = (int)in();
            long arr[] = arr(2*n);
           StringBuilder s = new StringBuilder();
          HashMap<Long,Integer> m = new HashMap<>();
           for(int j =0;j<2*n;j++){

               if(!m.containsKey(arr[j])){
                   s.append(arr[j]+" ");
                   m.put(arr[j],0);
               }
           }
            System.out.println(s.toString().trim());
        }
    }




    static long[] arr(long n) throws Exception {

        String read[] = sarrin();
        long arr[] = new long[(int) n];
        for (int k = 0; k < n; k++)
            arr[k] = in(read[k]);
        return arr;

    }

    static long[][] arr(long n, long m) throws Exception {


        long arr[][] = new long[(int) n][(int) m];
        for (int k = 0; k < n; k++) {
            String read[] = sarrin();
            for (int j = 0; j < m; j++)
                arr[k][j] = in(read[k]);
        }
        return arr;

    }

    static long in() throws Exception {
        return Long.parseLong(br.readLine());


    }

    static long in(String num) throws Exception {
        return Long.parseLong(num);


    }


    static String sin() throws Exception {

        return br.readLine();
    }

    static String[] sarrin() throws Exception {

        return br.readLine().split(" ");

    }

    static String[] regexSplit() throws Exception {
        return br.readLine().split(" ?(?<!\\G)((?<=[^\\p{Punct}])(?=\\p{Punct})|\\b) ?");
    }


}






