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.ArrayList;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class C623 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner obj=new Scanner(System.in);
		int t=obj.nextInt();
		while(t-->0)
		{
			int n=obj.nextInt();
			Set<Integer> set=new HashSet();
			Set<Integer> set2=new HashSet();
			int[] arr=new int[n];
			for(int i=0;i<n;i++)
			{
				arr[i]=obj.nextInt();
				set.add(arr[i]);
			}
			ArrayList<Integer> a=new ArrayList<>();
			for(int i=0;i<n;i++)
			{
				for(int j=arr[i]+1;j<=2*n;j++)
				{
					if(!set.contains(j)&&(!set2.contains(j)))
					{
						//System.out.println("**");
						set2.add(arr[i]);
						set2.add(j);
						a.add(arr[i]);
						a.add(j);
						set.remove(arr[i]);
						set.remove(j);
						break;
					}
				}
			}
			if(set2.size()==2*n)
			{
				for(int i=0;i<2*n;i++)
				{
					System.out.print(a.get(i)+" ");
				}
				System.out.println();
			}
			else
				System.out.println("-1");
		}

	}

}
