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 Solution{
    public static void main(String args[]){
        Scanner in=new Scanner (System.in);
        int t=in.nextInt();
        while(--t>=0){
            int n=in.nextInt();
            int a[]=new int[n];
            int b[]=new int[2*n];
            HashSet<Integer> hset=new HashSet<>();
            int flag=0;
            for(int i=0;i<n;i++){
                a[i]=in.nextInt();
                hset.add(a[i]);
                b[2*i]=a[i];
            }
                
            for(int i=0;i<n;i++){
                int x=a[i];
                while(true){
                    x++;
                    if(x>2*n){
                        flag=1;
                        break;
                    }
                    if(!hset.contains(x)){
                        hset.add(x);
                        b[2*i+1]=x;
                        break;
                    }
                }
            }
            if(flag==1)
                System.out.println(-1);
            else{
                for(int i=0;i<2*n;i++){
                    System.out.print(b[i]+" ");
                }
                System.out.println();
            }
        }
    }
}