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 Test2 {
     
        public static void main(String[] args) {           
            Scanner scanner = new Scanner(System.in);
            
            int test = scanner.nextInt();
            
            for(int t = 0; t<test; ++t){
                
                int n = scanner.nextInt();
                int[] sc = new int[(2 * n )+1];
                
                int[] op = new int[(2 * n )+1];
                int[] a = new int [n+1];
                for(int i=1; i<=n; ++i){
                    a[i] = scanner.nextInt();
                    
                    sc[a[i]] = 1;
                    op[2*i-1] = a[i];
                }
                
                boolean found = false;
                for(int i=1; i<=n; ++i){
                    
                    found = false;
                    for(int j = a[i]+1; j<=(2*n) ; ++j){
                        if(sc[j]==0){
                            op[2*i] = j;
                            sc[j] = 1;
                            found = true;
                            break;
                        }
                    }
                    if(!found){
                        break;
                    }
                    
                }
                
                if(!found){
                    System.out.println("-1");
                }
                else{
                    for(int i=1; i<=(2*n); ++i){
                        System.out.print(op[i] + " ");
                    }
                    System.out.println();
                }
                
                
                
            }
        }
             
    }