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 javax.jws.soap.SOAPBinding;
import java.io.*;
import java.lang.reflect.Array;
import java.security.acl.LastOwnerException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.stream.Collectors;


public class Main {

    static Long max = (long) (2 * 10E5);
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

    // Arrays.stream(br.readLine().split("\\s+")).map(Integer::parseInt).toArray(Integer[]::new);
    // Arrays.stream(br.readLine().split("\\s+")).map(Long::parseLong).toArray(Long[]::new);
    // Integer.parseInt(br.readLine());
    // Long.parseLong(br.readLine());


    public static void main(String[] args) throws IOException {
        int t = Integer.parseInt(br.readLine());
        for (int _t = 0; _t < t; _t++) {
            int n = Integer.parseInt(br.readLine());
            Integer[] b = Arrays.stream(br.readLine().split("\\s+")).map(Integer::parseInt).toArray(Integer[]::new);
            HashSet<Integer> original = new HashSet<>(Arrays.asList(b));
            HashSet<Integer> notUsed = new HashSet<>();

            for (int i = 1; i <= 2 * n; i++)
                if (!original.contains(i))
                    notUsed.add(i);

            ArrayList<Integer> permutation = new ArrayList<>();

            for (int i = 0; i < n; i++) {
                permutation.add(b[i]);
                for (int j = b[i] + 1; j <= 2 * n; j++)
                    if (notUsed.contains(j)) {
                        permutation.add(j);
                        notUsed.remove(j);
                        break;
                    }
                if (permutation.size() != (i + 1) * 2)
                    break;
            }
            if (permutation.size() == 2 * n){
                for (int i = 0; i < 2 * n - 1; i++)
                    bw.write(permutation.get(i) + " ");
                bw.write(permutation.get(2 * n - 1) + "\n");
            } else
                bw.write("-1\n");
        }
        bw.flush();
    }
}