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.io.*;

public class Solution
{
    static class Reader {
		BufferedReader br;
		StringTokenizer st;
 
		public Reader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		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;
		}
 
	}
    static class Ele implements Comparable<Ele> 
    {  
        public int x,y;  
        Ele(int x,int y) 
        {  
            this.x=x;this.y=y; 
        }  
        public int compareTo(Ele ob) {  
        if(ob.x!=x)return x-ob.x;
        return this.y-ob.y;  
        }   
    }
    int a[][];int b[][];
    int solve(int i,int j)// 0-none, 1-true,2-false
    {
        //System.out.println(i+" "+j);
        if (i==a.length-1 || j==a.length-1) return 1;
        else {
            if (b[i][j]!=0) return b[i][j];
            b[i][j]=2;
            if (a[i+1][j]==1) 
            b[i][j]=solve(i+1,j);
            if (a[i][j+1]==1) 
            b[i][j]=Math.min(b[i][j],solve(i,j+1));
            return b[i][j];
        }
    }
    long gcd(long a,long b)
    {
        long min=Math.min(a,b);
        long max=Math.max(a,b);
        while (max%min!=0)
        {
            a=max%min;
            max=min;min=a;
        }
        return min;
    }
	public static void main(String[] args) throws IOException 
	{
		Reader sc=new Reader();Solution G=new Solution();
		PrintWriter o = new PrintWriter(System.out);
		//int t=1;
		int t=sc.nextInt();
		
		int n,m,y,s1,s2;int a[],b[];long k,x;
		boolean b1,b2;String s;
		//long l;long a[]; 
		ArrayList<ArrayList<Integer>> ob=new ArrayList<>();
		ArrayList<Long> ob2=new ArrayList<>();
		ArrayList<Long> ob3=new ArrayList<>();
		TreeSet<Integer> h=new TreeSet<>();
		HashMap<Integer,Integer> h1=new HashMap<>();
		try{
		while (t-->0)
		{
		    n=sc.nextInt();a=new int[2*n];
		    for(int i=0;i<n*2;i++)
		    {
		        a[i]=sc.nextInt();
		        if (!h.contains(a[i]))
		        {
		            o.print(a[i]+" ");
		            h.add(a[i]);
		        }
		    }
		    //o.println(Arrays.toString(b));
		    
		    //o.println(h1);
		    o.println();
		    //o.println(ob3);
		    //o.println();
		    h.clear();ob.clear();ob2.clear();ob3.clear();//m.clear();
		}
		}
		catch (Throwable e)
		{
		    e.printStackTrace();
		}
		//o.println("HI");
		
        o.flush();
        o.close();
	}
} 