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 
//package com.kraken.cf.cf623;
/*
 * Author: Kraken
 * Created on 2/25/2020 at 3:50 AM
 */

import java.io.*;
import java.util.*;

public class C {

	public static void main(String[] args) {
		FastReader sc = new FastReader();
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] b = new int[n], x = new int[2 * n + 1];
			for (int i = 0; i < n; i++) {
				b[i] = sc.nextInt();
				x[b[i]]++;
			}
			StringBuilder res = new StringBuilder();
			boolean found = false;
			for (int i = 0; i < n; i++) {
				res.append(b[i]).append(" ");
				found = false;
				for (int j = b[i] + 1; j < 2 * n + 1; j++) {
					if (x[j] == 0 && j > b[i]) {
						x[j] = 1;
						res.append(j).append(" ");
						found = true;
						break;
					}
				}
				if (!found) {
					break;
				}
			}
			if (!found)
				System.out.println(-1);
			else
				System.out.println(res.toString());
		}
	}

	static class FastReader {

		BufferedReader br;

		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			}
			catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}
