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 
import java.io.*;
import java.util.*;
 
public class Main {
	static Scanner sc = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
 
	static void solve() throws Exception {
		int n = sc.nextInt(), arr[] = new int[2 * n];
		TreeSet<Integer> st = new TreeSet<>();
		for (int i = 1; i <= 2 * n; i++) {
			st.add(i);
		}
		for (int i = 0; i < n; i++) {
			arr[2 * i] = sc.nextInt();
			st.remove(arr[2 * i]);
		}
		for(int i = 1; i < 2*n; i+=2) {
			int val = arr[i - 1];
			if(st.higher(val) == null) {
				out.println(-1);
				return;
			}
			arr[i] = st.higher(val);
			st.remove(arr[i]);
		}
		for(int x : arr) out.print(x + " ");
		out.println();
	}
 
	public static void main(String[] args) throws Exception {
		int tc = sc.nextInt();
		while (tc-- > 0) {
			solve();
		}
		out.close();
	}
 
}
 
class Scanner {
	StringTokenizer st;
	BufferedReader br;
 
	public Scanner(InputStream system) {
		br = new BufferedReader(new InputStreamReader(system));
	}
 
	public Scanner(String file) throws Exception {
		br = new BufferedReader(new FileReader(file));
	}
 
	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
 
	public String nextLine() throws IOException {
		return br.readLine();
	}
 
	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
 
	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
 
	public Long nextLong() throws IOException {
		return Long.parseLong(next());
	}
}