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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Problem623C {
	public static void main(String[] args) throws Exception {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter writer = new PrintWriter(System.out);
		int testCases = Integer.parseInt(reader.readLine());
		for (int t = 0; t < testCases; t++) {
			int n = Integer.parseInt(reader.readLine());
			StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
			int[] b = new int[n];
			for (int i = 0; i < n; i++) {
				b[i] = Integer.parseInt(tokenizer.nextToken());
			}
			writer.println(solve(b));
		}
		writer.close();
	}
	public static String solve(int[] b) {
		StringBuilder builder = new StringBuilder();
		int n = b.length;
		boolean[] nums = new boolean[2 * n + 5];
		for (int i : b) {
			nums[i] = true;
		}
		for (int i = 0; i < b.length; i++) {
			builder.append(b[i]);
			builder.append(' ');
			boolean flag = false;
			for (int j = b[i] + 1; j <= 2 * n; j++) {
				if (!nums[j]) {
					builder.append(j);
					builder.append(' ');
					nums[j] = true;
					flag = true;
					break;
				}
			}
			if (!flag) {
				return "-1";
			}
		}
		return builder.toString().substring(0, builder.length() - 1);
	}
}