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.util.Scanner;
import java.util.ArrayList;

public class Main {
    public static void main(String[]args){
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        int size[] =new int[t];
        ArrayList list = new ArrayList();
        for(int i = 0; i < t;++i){
            int s = in.nextInt();
            int a[] = new int[s];
            size[i] = s;
            int m = 0;
            int arr[] = new int[2 * s];
            for(int j = 0; j < 2 * s;++j){
                arr[j] = in.nextInt();
                boolean r =true;
                for(int k = 0; k < s; ++k){
                    if(arr[j] == a[k])
                        r =false;
                }
                if(r){
                    a[m] = arr[j];
                    ++m;
                }
            }
            for(Object d: a){
                list.add(d);
            }
        }
        int d =0;
        for(int i = 0; i < t; ++i){
            for(int k = 0 ; k < size[i]; ++k){
                System.out.print(list.get(d) + "   ");
                ++d;
            }
            System.out.println();
        }
    }
}