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.*;
public class Main
{
    public static boolean isPresent(ArrayList<Integer> arr, int element){
        for(int i = 0; i<arr.size(); i++){
            if(arr.get(i)==element) return true;
        }
        return false;
    }
    
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0){
		    int n = sc.nextInt();
    		ArrayList<Integer> ar = new ArrayList<Integer>();
    		for(int i = 0; i<2*n; i++){
    		    ar.add(sc.nextInt());
    		}
    		
    		ArrayList<Integer> res1 = new ArrayList<Integer>();
    		ArrayList<Integer> res2 = new ArrayList<Integer>();
    		
    		for(int i = 0; i<2*n; i++){
    		    if(isPresent(res1,ar.get(i))) res2.add(ar.get(i));
    		    else res1.add(ar.get(i));
    		}
    	    for(int i = 0; i<n; i++){
    		    System.out.print(res1.get(i)+" ");
    		}
    		System.out.println();
    	}
	}
}