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 MyClass {
    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];
            Set<Integer> set=new HashSet<>();
            for(int i=0;i<n;i++)set.add(b[i]=sc.nextInt());
            int a[]=new int[2*n];
            int last=1;
            for(int i=0;i<n;i++){
                a[2*i]=b[i];
                int next=0;
                for(int j=b[i]+1;j<=2*n;j++){
                    if(!set.contains(j)){
                        set.add(j);
                        next=j;
                        break;
                    }
                }
                a[2*i+1]=next;
            }
            if(set.size()<2*n)
                System.out.println(-1);
            else{
                for(int i:a)System.out.print(i+" ");
                System.out.println();
            }
        }
    }
}