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.HashSet;
import java.util.Scanner;
import java.util.Set;

public class ProblemC {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int tt = 0; tt < t; tt++) {
            int n = sc.nextInt();
            int[] input = new int[n];
            int[] output = new int[2*n];
            Set<Integer> used = new HashSet<>();
            for(int i = 0; i < n; i++) {
                input[i] = sc.nextInt();
                used.add(input[i]);
            }
            int max = 2*n;
            boolean impossible = false;
            for(int i = 1; i < 2*n; i += 2) {
                output[i-1] = input[i/2];
                int nextNotUsed = output[i-1]+1;
                while (used.contains(nextNotUsed)) nextNotUsed++;
                if(nextNotUsed > max) {
                    impossible = true;
                    break;
                }
                used.add(nextNotUsed);
                output[i] = nextNotUsed;
            }

            if(impossible) {
                System.out.println(-1);
            } else {
                for(int i = 0; i < 2* n; i++) {
                    System.out.print(output[i]);
                    if(i < (2 * n) -1) {
                        System.out.print(" ");
                    }
                }
                System.out.println();
            }
        }
    }
}