Problem: 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 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.math.*;


public class Codeforces {

    public static void main(String[] args) throws IOException {
        Scanner s = new Scanner(System.in);
        int t=s.nextInt();
        for(int i=0;i<t;i++)
        {
           int n=s.nextInt();
           int[] arr=new int[n];
           HashSet<Integer> set=new HashSet<>();
           for(int j=0;j<n;j++)
           {
               arr[j]=s.nextInt();
               set.add(arr[j]);
           }
           int flag=0;
           int[] ans=new int[2*n];
           for(int j=0;j<n;j++)
           {
               int val=arr[j];

               int minReq=-1;
               for(int k=val+1;k<=2*n;k++)
               {
                   if(!set.contains(k))
                   {
                       minReq=k;
                       break;
                   }
               }
//               System.out.println(val+" "+minReq);
               if(minReq==-1)
               {
                   flag=1;
                   break;
               }
               {
                   ans[2*j]=arr[j];
                   ans[(2*j)+1]=minReq;
                   set.add(minReq);
               }
           }
           if(flag==1)
           {
//               for(int j:ans)
//               {
//                   System.out.print(j+" ");
//               }
               System.out.println("-1");
           }
           else
           {
               for(int j:ans)
               {
                   System.out.print(j+" ");
               }
               System.out.println();
           }

        }


    }
}