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.Scanner;


public class JavaApplication2 {

    public static void main(String[] args) {
        Scanner s=new Scanner(System.in);
        int t=s.nextInt();
        String ss="";
      int counter=0;
        
        for(int i=0;i<t;i++){
            int n=s.nextInt();
            boolean doo=true;
            
            int[]ar=new int[n];
           
            int []an=new int[2*n];
            for(int o=0;o<an.length;o++)
                an[o]=o+1;
            
            for(int x=0;x<n;x++){
                ar[x]=s.nextInt();
                if(ar[x]>=2*n)
                    doo=false;
                    
            }
            if(doo){
                for(int is=0;is<ar.length;is++){
                    for(int c=0;c<an.length;c++){
                        if(ar[is]<an[c]&&(exist(ar,an[c]))){
                            ss+=ar[is]+" "+an[c]+" ";
                            an[c]=0;
                            counter++;
                            break;
                         
                        }
                     
                      
                    }
                    
                }
                if(counter<n){
                    System.out.println(-1);
                    ss="";
                    
                }
                else{
                    System.out.println(ss);
                    ss="";
                }
                
         
            counter=0;
        
            }
               
            else{System.out.println(-1);}
        }
            
        }
        
        
    
    public static boolean exist(int ar[],int s){
        boolean x=true;
        int gg;
        for(int d=0;d<ar.length;d++)
            if(ar[d]==s)
                x=false;
        return x;
    }
    
}