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.Arrays;
import java.util.Scanner;
import java.util.Stack;

import javax.sound.sampled.ReverbType;

import java.util.ArrayList;
import java.util.List;
import java.lang.reflect.Array;
import java.math.*;

public class Sec4 {

	public static void main(String[] args) {

		Scanner read = new Scanner(System.in);
		int t = read.nextInt();
		for(int k=0;k<t;k++)
		{
		ArrayList<Integer>b=new ArrayList<Integer>();
		int n = read.nextInt();
		int  a []=new int [2*n];
		for (int i=0;i<2*n;i++)
		{
			a[i]=read.nextInt();
		}
		int g = 0;
		for (int i=0;i<2*n;i++)
		{
			if (i==0)
			{
				b.add(a[i]);
			}
			else {
				boolean b1 =false;
			for(int j=0;j<b.size();j++)
			{
				if (a[i]==b.get(j))
				{
					b1=true;
				}
			}
			if (b1==false)
			{
				b.add(a[i]);
			}
			
			}
		}
		for (int i = 0 ;i<b.size();i++) {
		System.out.print(b.get(i)+" ");
		}
		System.out.println("");
		}
	}
	
	



	
}