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


public class q_2 {
    static Scanner sc;
    static FastReader fr;
    public static void main(String[] args){
        sc = new Scanner(System.in);
        fr = new FastReader();
        int test = fr.nextInt();
        for(int tt=0; tt<test; tt++){
            int n = fr.nextInt();
            Integer[] arr = getIntArrayFromSingleLine();
            HashSet<Integer> numSet = new HashSet<>(10*n);
            StringBuilder sb = new StringBuilder();
            for(int i=0; i<2*n; i++){
                if(numSet.contains(arr[i])){
                    // do nothing
                }
                else {
                    numSet.add(arr[i]);
                    //System.out.println("Appending num "+arr[i]);
                    sb.append(arr[i]+" ");
                }
            }
            System.out.println(sb.toString());
        }
    }

    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());
        }

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

    public static Integer[] getIntArrayFromSingleLine(){
        List<Integer> list = new LinkedList<>();
        String[] str = fr.nextLine().split(" ");
        for(String s: str){
            list.add(Integer.parseInt(s));
        }
        Integer[] arr = new Integer[list.size()];
        return list.toArray(arr);
    }

    public static Long[] getLongArrayFromSingleLine(){
        List<Long> list = new LinkedList<>();
        String[] str = fr.nextLine().split(" ");
        for(String s: str){
            list.add(Long.parseLong(s));
        }
        Long[] longs = new Long[list.size()];
        return list.toArray(longs);
    }

    public static Double[] getDoubleArrayFromSingleLine(){
        List<Double> list = new LinkedList<>();
        String[] str= fr.nextLine().split(" ");
        for(String s: str){
            list.add(Double.parseDouble(s));
        }
        Double[] doubles = new Double[list.size()];
        return list.toArray(doubles);
    }

    public static String[] getStringArrayFromSingleLine(){
        return fr.nextLine().split(" ");
    }

    public static int[] getArray(int n){
        int[] arr = new int[n];
        for(int i=0; i<n; i++){
            arr[i] = fr.nextInt();
        }
        return arr;
    }

    public static long[] getLongArray(int n){
        long[] res = new long[n];
        for(int i=0; i<n; i++){
            res[i] = fr.nextLong();
        }
        return res;
    }

    public static double[] getDoubleArray(int n){
        double[] arr = new double[n];
        for(int i=0; i<n ; i++){
            arr[i] = fr.nextDouble();
        }
        return arr;
    }

    public static char[] getCharArray(){
        return fr.next().toCharArray();
    }

    public static String[] getStringArray(int n){
        String[] strings = new String[n];
        for(int i=0; i<n; i++){
            strings[i] = fr.next();
        }
        return strings;
    }

    // Get GCD
    public static int getGCD(int a, int b){
        return (a==0 ? b : getGCD(b%a, a));
    }

    public static int getGCD(int[] arr){
        int n = arr.length;
        if(n==0){
            return -1;
        }
        int res = arr[0];
        for(int i=1; i<n; i++){
            res = getGCD(res, arr[i]);
        }
        return res;
    }

    // Get all prime numbers till N using seive of Eratosthenes
    public static List<Integer> getPrimesTill(int n){
        boolean[] arr = new boolean[n+1];
        List<Integer> primes = new LinkedList<>();
        for(int i=2; i<=n; i++){
            if(!arr[i]){
                primes.add(i);
                for(long j=(i*i); j<=n; j+=i){
                    arr[(int)j] = true;
                }
            }
        }
        return primes;
    }


    // This method returns list of PrimeFactors of a number in O(sqrt(N));
    public static Map<Integer,Integer> getPrimeFactors(long n){
        Map<Integer,Integer> res = new HashMap<>();
        int sqrt = (int) Math.sqrt(n);
        if(n%2 == 0){
            int c = 0;
            while(n%2==0){
                c++;
                n/=2;
            }
            res.put(2,c);
        }
        for(int i=3; i<=sqrt; i+=2){
            if(n%i == 0) {
                int c = 0;
                while (n % i == 0) {
                    c++;
                    n /= i;
                }
                res.put(i,c);
            }
        }
        if(n!=1){
            res.put((int)n,1);
        }
        return res;
    }

}