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.util.*;

public class Main { 

	static void println(Object o){
		System.out.println(o);
	}

	public static void main(String[] args){
		
		FastScanner sc = new FastScanner();
		int t = sc.nextInt(), n = 0, temp = 0;
		while(t-- > 0){
			n = sc.nextInt();
			HashSet<Integer> s = new HashSet<>();
			ArrayList<Integer> al = new ArrayList<>();
			for(int i = 0; i<2*n; i+=1){
				temp = sc.nextInt();
				if(!s.contains(temp)){
					s.add(temp);
					al.add(temp);
				}
			}
			StringBuilder res = new StringBuilder();
			for(int i = 0; i<n; i+=1){
				res.append(al.get(i));
				res.append(' ');
			}
			println(res);
		}
		
	}
}

class FastScanner {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = null;

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

		String next(){
			while(st == null || !st.hasMoreTokens()){
				try{
					st = new StringTokenizer(br.readLine());
				}
				catch(IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt(){
		    return Integer.parseInt(this.next());
		}
		
		long nextLong(){
		    return Long.parseLong(this.next());
		}

}