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 
/*
  Created by: Weslley (w3s)
*/

import java.util.*;
import java.io.*;
import java.util.StringTokenizer;


public class A1385{
    static FastScanner fs;
    
    public static void solve(){
	ArrayList<Integer> set = new ArrayList<>();
	int n = fs.nextInt();
	for(int i=0;i<2*n;i++){
	    int x = fs.nextInt();
	    if(!set.contains(x))
		set.add(x);
	}
	for(Integer x : set){
	    System.out.print(x+" ");
	}
	System.out.println();
    }
    
    public static void main(String[] args){
	fs = new FastScanner(System.in);
	int t  = fs.nextInt();
	while(t-->0){
	    solve();
	}
	FastPrint.out.close();
    }
}

class FastScanner {
    BufferedReader br;
    StringTokenizer st;
 
    public FastScanner(InputStream stream) {
	br = new BufferedReader(new InputStreamReader(stream));
    }
 
    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;
    }

}

class FastPrint {
    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
}





