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.math.BigInteger;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class Main {
    public static void main(String[] args) throws IOException {
        int t = nI();
        for (int i = 0; i < t; i++) {
            int len = nI();
            int[] perm = new int[len*2];
            for (int j = 0; j < len*2; j++) {
                perm[j] = nI();
            }
            HashSet<Integer> hs = new HashSet<>();
            int[] ans = new int[len];
            int m = 0;
            for(Integer j: perm){
                if(!hs.contains(j)){
                    hs.add(j);
                    ans[m] = j;
                    ++m;
                }
            }
            for (Integer j: ans) {
                out.print(j+" ");
            }
            out.println();
        }
        out.close();
    }


































    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    //static BufferedReader brfile;
    //static PrintWriter outfile;
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer in = new StringTokenizer("");

   /* static {
        try {
            brfile = new BufferedReader(new FileReader("input.txt"));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }

    static {
        try {
            outfile = new PrintWriter("output.txt");
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }

*/
    public static String nT() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public static int nI() throws IOException {
        return Integer.parseInt(nT());
    }

    public static double nD() throws IOException {
        return Double.parseDouble(nT());
    }

    public static long nL() throws IOException {
        return Long.parseLong(nT());
    }

    public static boolean nB() throws IOException {
        return Boolean.parseBoolean(nT());
    }
}