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.*;
import java.io.*;
public class C623
{
    public static void main(String [] args)
    {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = sc.nextInt();
        while (t > 0) {
            int n = sc.nextInt();
            int [] res = new int [2*n];
            TreeSet <Integer> sorted = new TreeSet <Integer> ();
            for (int i = 1; i <= 2*n; i++) {
                sorted.add(i);
            }
            for (int i = 0; i < n; i++) {
                res[2*i] = sc.nextInt();
                sorted.remove(res[2*i]);
            }
            for (int i = 0; i < n; i++) {
                Integer ideal = sorted.ceiling(res[2*i]);
                if (ideal == null) {

                } else {
                    res[2*i + 1] = ideal;
                    sorted.remove(ideal);
                }
            }

            boolean bad = false;
            for (int i = 0; i < res.length; i++) {
                if (res[i] == 0) {
                    bad = true;
                    break;
                }
            }

            if (bad) {
                out.println(-1);
            } else {
                for (int i = 0; i < res.length; i++) {
                    out.print(res[i] + " ");
                }
                out.println();
            }



            t--;
        }

        out.close();
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }



    }

}