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.*;
import java.util.*;
public class RestoringPermutation {
	private static StreamTokenizer st;
	private static int nextInt()throws IOException{
		st.nextToken();
		return (int)st.nval;
	}
	public static void main(String[] args) throws IOException{
		st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		int t = nextInt();
		TreeSet<Integer> ts = new TreeSet<>();
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		for(int x = 0; x < t; ++x) {
			int n = nextInt();
			boolean used[] = new boolean[2*n];
			int b[] = new int[n];
			for(int i = 0; i < n; ++i) {
				b[i] = nextInt()-1;
				used[b[i]] = true;
			}
			ts.clear();
			for(int i = 0; i < 2*n; ++i) if(!used[i])ts.add(i);
			int a[] = new int[2*n];
			boolean valid = true;
			for(int i = 0; i < n; ++i) {
				a[i*2] = b[i]+1;
				if(ts.higher(b[i]) == null) {
					valid = false;
					break;
				}
				int ret = ts.higher(b[i]);
				a[i*2+1] = ret+1;
				ts.remove(ret);
			}
			if(!valid)pw.println(-1);
			else {
				for(int i = 0; i < 2*n-1; ++i)pw.print(a[i] +" ");
				pw.println(a[2*n-1]);
			}
		}
		pw.close();
	}

}