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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Comparator;
import java.util.Scanner;
import java.util.Arrays;
import java.util.TreeSet;
import java.util.StringTokenizer;


public class vk18
{
    public static void main(String[] stp) throws Exception
    {
        Scanner scan = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int t=Integer.parseInt(st.nextToken()),i,j=0;
        while(t-->0)
        {
            st=new StringTokenizer(br.readLine());
            int n=Integer.parseInt(st.nextToken()),min=Integer.MAX_VALUE;
            Pair pair[]=new Pair[n];
            boolean flag[]=new boolean[2*n + 1];
            Arrays.fill(flag,false);
            st = new StringTokenizer(br.readLine());
            for(i=0;i<n;i++)
            {
                int p=Integer.parseInt(st.nextToken());
                pair[i]=new Pair(p,i+1,0);
                flag[p]=true;
                min=Math.min(min,p);
            }
            TreeSet<Integer> ts=new TreeSet<>();
            for(i=1;i<=2*n;i++) { if(flag[i]==false) { ts.add(i);  } }
            //Arrays.sort(pair,new Pair.First());
            boolean right=true,left=true;
            for(i=0;i<n;i++)
            {
                right=false;
                int find=pair[i].d;
                for(j=pair[i].d+1;j<=2*n;j++)
                {
                    if(ts.contains(j)) { right=true; pair[i].e=j; ts.remove(j); break; }
                }
                if(right==false){ pw.print(-1); left=false; break; }

            }
            if(left)
            {
                //Arrays.sort(pair, new Pair.Second());
                for (i = 0; i < n; i++) pw.print(pair[i].d + " " + pair[i].e+" ");
            }
            pw.println();
        }
        pw.flush();

    }
    static class Pair implements Comparable<Pair>
    {
        int d, g,e;

        public Pair(int d, int g,int e)
        {
            this.d = d;
            this.g = g;
            this.e = e;
        }

        @Override
        public int compareTo(Pair p) {
            return 1;
        }

        static class First implements Comparator<Pair>
        {
            public int compare(Pair q, Pair s)
            {
                return -Integer.compare(s.d, q.d);
            }
        }

        static class Second implements Comparator<Pair>
        {
            public int compare(Pair q, Pair s)
            {
                return -Integer.compare(s.g, q.g);
            }
        }
        static class Third implements Comparator<Pair>
        {
            public int compare(Pair q, Pair s)
            {
                return -Integer.compare(s.e, q.e);
            }
        }
    }

}