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.util.*;

 public class Sol{
   public static void main(String[] args){
  Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    while(t-->0){ 
       int n=sc.nextInt();
     int[] b=new int[n+1];
     int[] bb=new int[n];
     int le=2*n;
     Set<Integer> h=new HashSet<>(); int min=0;
     ArrayList<Integer> al= new ArrayList<>();
     ArrayList<Integer> alb= new ArrayList<>();
      ArrayList<Integer> aa= new ArrayList<>();
      Map<Integer,Integer> mp=new HashMap<>();
     for(int i=1;i<=n;i++){ b[i]=sc.nextInt();
     bb[i-1]=b[i];
         alb.add(b[i]);
         
         h.add(b[i]);
         if(min<b[i]){
             min=b[i];
         }
         
        
     }
    
     boolean ok=true;
     
    
      for(int i=1;i<=n ;i++){
          ok=false;
          
               for(int jjj=0;jjj<=le;jjj++){
           int j=b[i]+jjj;
          if(!alb.contains(j)&& ok!=true && j<=le){
              alb.add(j);
              aa.add(j);
              ok=true;
              
          }
          
        }
      }
      
     /* int p=1;
     Arrays.sort(bb);
     Collections.sort(aa);
     int[] cc=new int[le+1];
     for(int i=1;i<=le;i+=2){
         
        
         mp.put( bb[i-p],aa.get(i-p));
         p++;
         
         
     }
     */
    
     
     if( h.size()!=n|| min>=n*2 ||aa.size()!=n){
         System.out.println("-1");
     }
     else{
         for(int i=1;i<=n;i++){
              
             
            System.out.print( b[i]+" "+aa.get(i-1)+" ");
             
         }
         System.out.println();
     }
        
        
    
    
}
}
}