You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).

The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).

The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).

It is guaranteed that the sum of n by all test cases doesn't exceed 100.

Output

For each test case, if there is no appropriate permutation, print one number -1.

Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.

Example

Input


5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8


Output


1 2 
-1
4 5 1 2 3 6 
-1
1 3 5 6 7 9 2 4 8 10 
import java.util.*;

public class RestoringPermutation{
    public static int find(Set<Integer> unusedNums, int a)
    {
        int min = 300; 
        for(int i : unusedNums)
        {
            if(i > a && i < min)
                min = i;
        }
        return min;
    }
    public static void main(String []args){
         Scanner s1 = new Scanner(System.in);
         int t = Integer.parseInt(s1.nextLine());
         for(int q = 0; q < t; q++)
         {
             int n = Integer.parseInt(s1.nextLine());
             String s = s1.nextLine();
             String[] sequence = s.split(" ");
             int[] arr = new int[2*sequence.length];
             Set<Integer> usedNums = new HashSet<Integer>(); 
             Set<Integer> unusedNums = new HashSet<Integer>(); 
             for(int i = 0; i < sequence.length; i++)
            {    
                arr[2*i] = Integer.parseInt(sequence[i]);
                usedNums.add(Integer.parseInt(sequence[i]));
            }
             
             for(int i = 1; i <= 2*sequence.length; i++)
             {
                if(!usedNums.contains(i))
                    unusedNums.add(i);
             }
             
             for(int i = 0; i < sequence.length; i++)
            {    
                arr[2*i + 1] = find(unusedNums, arr[2*i]);
                if(arr[2*i+1] == 300)
                {
                    System.out.println(-1);
                    break;
                }
                unusedNums.remove(arr[2*i+1]);
            }
            if(unusedNums.size() ==0){
            for(int i = 1; i <= 2*sequence.length; i++)
             {
                 System.out.print(arr[i-1] + " ");
             }
             System.out.println();}
         }
     }
}