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 Solution{
	long rem = 1000000007L;
	public Solution(){
		Scanner sc = new Scanner(System.in);
		int tests = sc.nextInt();
        
        for(int t=1; t<=tests; t++){
			int n = sc.nextInt();
			int[] b = new int[n];
			boolean[] found = new boolean[(2*n)+1];
			boolean possible = false;
			
			for(int i=0; i<n; i++){
				b[i] = sc.nextInt();
				found[b[i]] = true;
			}
			int[] res = new int[2*n];
			int index = 0;
			
			for(int i=0; i<n; i++){
				res[index++] = b[i];
				possible = false;
				for(int j=b[i]+1; j<found.length; j++){
					if(!found[j]){
						found[j] = true;
						res[index++] = j;
						possible = true;
						break;
					}
				}
				if(!possible){
					break;
				}
			}
			if(!possible){
				System.out.println("-1");
			}
			else{
				for(int i=0; i<(res.length-1); i++){
					System.out.print(res[i] + " ");
				}
				System.out.println(res[res.length-1]);
			}
        }
	}
	
	public static void main(String[] args){
		new Solution();
	}
}