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.Scanner;
public class main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        Counter ct = new Counter();

        int test_count = sc.nextInt();
        for (int i = 0; i < test_count; i++) {
            int size = sc.nextInt();
            int[] mass = new int[size*2];
            for (int j = 0; j <size*2 ; j++) {
                mass[j] = sc.nextInt();
            }
            mass = ct.setData(size,mass);
            for (int j = 0; j <size ; j++) {
                System.out.print(mass[j]+" ");
            }
            System.out.println();
        }
    }
}

class Counter {
    int fusedData[];
    int data[];

    int[] setData(int size,int[] fusedData_) {
        //long start = System.nanoTime();
        fusedData = fusedData_;
        data = new int[size];
        int pos = 0;
        for (int i = 0; i < size * 2; i++) {
            boolean flag = false;
            for (int k : data) {
                if (k == fusedData[i]) flag = true;
            }
            if(!flag) {
                data[pos] = fusedData[i];
                pos++;
            };
        }
        //long finish = System.nanoTime();
        //long timeConsumedMillis = finish - start;
        //System.out.println("Время выполнения: " + timeConsumedMillis);
        return data;
    }


}

