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.BufferedReader;
	import java.io.IOException;
	import java.io.InputStreamReader;
	import java.math.BigInteger;
	import java.util.ArrayList;
	import java.util.Arrays;
	import java.util.Collections;
	import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
	import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
	
	
	public class Main {
		   static class FastReader 
		    { 
	       BufferedReader br; 
	       StringTokenizer st; 
	      
	 
	       public FastReader() 
	       { 
	           br = new BufferedReader(new
	                    InputStreamReader(System.in)); 
	       } 
	 
	       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(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; 
	       }
	
			public Character charAt(int i) {
				// TODO Auto-generated method stub
				return null;
			}
	
			public BigInteger nextBigInteger() {
				// TODO Auto-generated method stub
				return null;
			} 
	   } 	
	
	//	      int a = 3125, m = 9987; 
		 //   System.out.println( modInverse(a, m));
	//	 	GIVE BIT AT A PARTICULAR POSITION 
	//			if((n&(1<<pos))!=0) System.out.println(1);
	//			else System.out.println(0);
	//		FOR SET
	//			System.out.println((n|(1<<pos))); 
	//		FOR CLEAR 
	//			System.out.println((n & ~(1<<pos)));
		//  FOR UPDATE
	//		 int k = (n  &~(1<<t)); CLEAR
	//		 System.out.println((k | (1<<t))); SET	   
		   
	//	   int t = s.nextInt();
	//		while(t-->0) {
	//			int n = s.nextInt();
	//		}	
		  
			public static void main(String[] args) {
				FastReader s=new FastReader();
				int t = s.nextInt();
				while(t-->0) {
					int n = s.nextInt();
					List<Integer> se1 = new ArrayList<>();
					for(int i=0;i<2*n;i++) {
						int p=s.nextInt();
						if(!se1.contains(p)) {
							se1.add(p);
//							System.out.println(p);
						}
//						System.out.println(i);
					}
					for(int i=0;i<n;i++) {
						System.out.print(se1.get(i)+" ");
					}
					System.out.println();
					
					
				}
			}			
		}
		 
		
	//
	//}
			
		
			