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 com.sun.jdi.ArrayReference;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
    public static void main(String[] args) {
//        Scanner sc = new Scanner(System.in);
       FastScanner sc = new FastScanner();
       int t=sc.nextInt();
       outer:while (t-->=1) {
           int n = sc.nextInt();
           int a[]=sc.readArray(2*n);
           ArrayList<Integer> ar= new ArrayList<>();
           
           for (int i=0;i<2*n;i++){

               if (ar.contains(a[i])){
                   continue;
               }
               else{
                   System.out.print(a[i]+" ");
                   ar.add(a[i]);
               }
           }
           System.out.println();


           }

       }

        static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }
    static void sortLong(long[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (long i : a) l.add((int)i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    static void sortReverse(long[] a) {
        ArrayList<Long> l = new ArrayList<>();
        for (long i : a) l.add(i);
        Collections.sort(l, Collections.reverseOrder());
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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


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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }
        long[] readArrayLong(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

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



