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;

public class C_623 {

	@SuppressWarnings("resource")
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int t = input.nextInt();
		for (int test = 0; test < t; test++){
			int n = input.nextInt();
			int[] b = new int[n + 1];
			boolean[] taken = new boolean[n * 2 + 1];
			for (int i = 1; i <= n; i++){
				b[i] = input.nextInt();
				taken[b[i]] = true;
			}
			String out = "";
			boolean no = false;
			for (int i = 1; i <= n; i++){
				out += b[i] + " ";
				int j = b[i] + 1;
				while (j <= n * 2 && taken[j]){
					j++;
				}
				if (j > n * 2){
					no = true;
					break;
				}
				taken[j] = true;
				out += j + " ";
			}
			//System.out.println(out);
			if(no){
				System.out.println(-1);
			}else{
				System.out.println(out.substring(0, out.length() - 1));
			}
		}
	}

}