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.Scanner;

//https://codeforces.com/problemset/problem/1315/C
public class ResortingPermutations {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            int[] arr = new int[n];
            for (int j = 0; j < n; j++) {
                arr[j] = sc.nextInt();
            }
            System.out.println(permutation(arr));

        }
    }

    private static String permutation(int[] ar) {
        int n = ar.length;
        int[] per = new int[(2 * n) + 1];
        String output = "";


        for (int i = 0; i < n; i++) {
            per[ar[i]] = 1;
        }
        for (int i = 0; i < n; i++) {
            boolean perm = false;
            final int b = ar[i];
            int j = b + 1;
            for (; j <= 2 * n; j++) {

                if (per[j] != 1) {
                    output = output + " " + b + " " + j;
                    per[j] = 1;
                    perm = true;
                    break;
                }
            }
            if (!perm) {
                int[] t = {};
                return output = "-1";
            }
        }
        return output;
    }
}