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.*;

public class prac1 {

	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		int i;
		for(i=0;i<t;i++) {
			int n = in.nextInt();
			int j=0;
//			int[] a = new int[n*2];
			Vector a = new Vector();
			for(j=0;j<(2*n);j++) {
				int nu =in.nextInt();
				a.add(nu);
			}
			
			int k=0;
			Vector b = new Vector();
			Vector c = new Vector();
//			int[] c = new int[50];
			int bi=0,lar=0,ci=0;
			int halfb=0,halfc=0;
			for(k=0;k<(2*n);k++) {
//			if(halfb<=((n)-1)&&halfc<=((n)-1)) {	
				if(k==0) {
					int as= (int) a.get(k);
					b.add(as);
					lar=(int)b.get(k);
					bi++;
					halfb=1;
				}
				else{
//					System.out.println(a[k]+" "+ k+"good");
					int num = (int)a.get(k);
//					Vector v = new Vector(); 
//					 System.out.println("index of Geeks is: " + v.indexOf("Geeks")); 
					int conta = b.indexOf(num);
					
//					System.out.println(conta+" "+num+" "+lar+"cont");
					if(num>lar) {
//						b[bi]=num;
//						System.out.println(num+" "+lar+"lar");
						b.add(num);
						bi++;
						lar=num;
						halfb++;
					}
					if((num<lar)&&(conta<0)) {
//						System.out.println(conta+"conta"+num);
						b.add(num);
						bi++;
//						lar=num;
						halfb++;
					}
					else {
//						c[ci]=num;
						c.add(num);
						ci++;
						halfc++;
					}
//					if(lar<num) lar=num;
					
//				}
				
			}
			}
			int y=0;
//			System.out.print(b);
			
			for(y=0;y<n;y++) {
				System.out.print(b.get(y)+" ");
			}
			System.out.println();

			
		
			
			
			
			
			
			
			
			
		}
	}

}


















//1
//10
//1
//1
//6
//6
//7
//7
//8
//8
//9
//9
//10
//5
//10
//5
//4
//3
//4
//3
//2
//2