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.*;
public class Solutions {
	static Scanner scr=new Scanner(System.in);
	static String s[];
	static String s1[];
	static int n;
	static int m;
	public static void main(String[] args) {
		int t=scr.nextInt();
		while(t-->0) {
			solve();
		}
	}
	static void solve() {
		int n=read();
		int []a=new int[2*n];
		for(int i=0;i<2*n;i++) {
			a[i]=read();
		}
		boolean visited[]=new boolean[n+1];
		for(int i=0;i<2*n;i++) {
			if(visited[a[i]]==false) {
				print(a[i]);
				visited[a[i]]=true;
			}
		}
		println();
	}
	static void swap(int a[],int i,int j) {
		int temp=a[i];
		a[i]=a[j];
		a[j]=temp;
	}
	static int len(String s) {
		return s.length();
	}
	static long abs(long a) {
		return Math.abs(a);
	}static int abs(int a) {
		return Math.abs(a);
	}
	static long min(long a,long b) {
		return Math.min(a, b);
	}static long max(long a,long b) {
		return Math.max(a, b);
	}
	static void print(int a) {
		System.out.print(a+" ");
	}static void print(long a) {
		System.out.print(a+" ");
	}static void print(String a) {
		System.out.print(a);
	}static void println(int a) {
		System.out.println(a);
	}static void println(long a) {
		System.out.println(a);
	}static void println(String a) {
		System.out.println(a);
	}
	static void println() {
		System.out.println();
	}
	static int read() {
		return scr.nextInt();
	}
	static String readS() {
		return scr.next();
	}
	static long readL() {
		return scr.nextLong();
	}
	static int[]readIntArray(int n){
		int []a=new int[n];
		for(int i=0;i<n;i++) {
			a[i]=scr.nextInt();
		}
		return a;
	}static long[]readLongArray(int n){
		long []a=new long[n];
		for(int i=0;i<n;i++) {
			a[i]=scr.nextLong();
		}
		return a;
	}
}
