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.*;
import java.io.*;

public class B{
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = fs.nextInt();
		m:for(int tt=0;tt<t;tt++)
		{
			int n = fs.nextInt();
			Set<Integer> s = new HashSet();
			int[] arr = new int[n];
			for(int i=0;i<n;i++)
			{
				int a = fs.nextInt();
				arr[i] = a;
				if(s.contains(a)==true)
				{
					System.out.println(-1);
					continue m;
				}
				s.add(a);
			}
			int[] ans = new int[2*n];
			boolean flag = true;
			int index = 0;
			for(int i=0;i<n;i++)
			{
				int a = arr[i];
				int need = a+1;
				while(s.contains(need)==true)need++;
				s.add(need);
				ans[index] = a;
				ans[index+1] = need;
				index+=2;
			}
			for(int i:s)
			{
				if(i>2*n)
				{
					flag = false;
					break;
				}
			}
			if(flag == false)
			{
				out.println(-1);
			}
			else
			{
				for(int i:ans)
				{
					out.print(i+" ");
				}
				out.println();
			}
		}
		out.close();
	}
	
		
		
	




	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

	
}