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.*;
import java.util.*;

public class Restoring_Permutation
{
    public static void main(String[] args) {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            int T = Integer.parseInt(br.readLine());

            for (int i = 1; i <= T; i++) {
                int N = Integer.parseInt(br.readLine());
                int[] arr = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
                check(arr);
            }
        } catch (Exception e) {
            return;
        }
    }

    private static void check(int[] arr)
    {
        HashSet<Integer> seq = new HashSet<>();
        for(int i=0;i<arr.length;i++){
            seq.add(arr[i]);
        }
        String s="";
        for(int i=0;i<arr.length;i++)
        {
            int k = arr[i] + 1;
            boolean found = false;
            while(k <= 2*arr.length)
            {
                if(!seq.contains(k)){
                    s += arr[i] + " " + k + " ";
                    found = true;
                    seq.add(k);
                    break;
                }
                k++;
            }
            if(!found){
                System.out.println("-1");
                return;
            }
        }
        System.out.println(s);
    }
}