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.util.*;
import java.util.Scanner;

public class RestorePermutation {

	static int[] solve(int[] in){
		HashMap<Integer,Integer> map = new HashMap<Integer, Integer>();
		for(int i=0;i<in.length;i++){
			map.put(in[i],i);
		}
		ArrayList<Integer>out = new ArrayList<Integer>();
		HashSet<Integer>processed = new HashSet<Integer>();
		int[]iarr=new int[in.length];
		
		for(int i=0;i<iarr.length;i++){
			iarr[i]=-1;
			
		}
		int i =0;
		int index=0;
		while(i < in.length){
			if(iarr[i] != -1){
				i++;
				continue;
			}
			int fi = i;
			int si = map.get(in[i]);
			iarr[si]=index;
			iarr[fi]=index;
			index++;
			i++;
		}
		int[]res = new int[in.length/2];
		index=0;
		for(int k=0;k<iarr.length;k++){
			int ele = in[k];
			int ie = iarr[k];
			res[ie]=ele;
		}
		return res;
	}
	public static void main(String[] args) {
		
		Scanner in = new Scanner(System.in);
		int tc = Integer.parseInt(in.nextLine());
		
		int n=1;
		
		while(n <= tc){
			int esize  = Integer.parseInt(in.nextLine());
			String elements = in.nextLine();
			String[]earr = elements.split("\\s+");
			int[]arr = new int[earr.length];
			for(int i=0;i<earr.length;i++){
				arr[i]=Integer.parseInt(earr[i]);
			}
			int[]out=solve(arr);
			for(int i=0;i<out.length;i++){
				System.out.print(out[i]+" ");
			}
			System.out.println();
			n++;
		}
	}
}

