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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author ky112233
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] b = new int[n];
            int[] arr = new int[2 * n];
            for (int i = 0; i < n; i++) {
                int num = in.nextInt();
                b[i] = num;
                arr[num - 1] = 1;
            }
            int[] comp = new int[n];
            for (int i = 0; i < n; i++) {
                int temp = b[i];
                int j;
                for (j = temp; j < 2 * n; j++) {
                    if (arr[j] == 0) {
                        arr[j] = 1;
                        comp[i] = j + 1;
                        break;
                    }
                }
                if (j == 2 * n) {
                    out.println(-1);
                    return;
                }
            }
            for (int i = 0; i < n; i++) {
                out.print(b[i] + " " + comp[i] + " ");
            }
            out.println();
        }

    }
}

