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 GFG {
    public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(br.readLine().trim());
		while(t--!=0){
			int n=Integer.parseInt(br.readLine().trim());
    		String s=br.readLine().trim();
    		String ss[]=s.split("\\s+");
			int a[]=new int[n];
			boolean vis[]=new boolean[2*n+1];
			for(int i=0;i<n;i++){
				a[i]=Integer.parseInt(ss[i]);
				vis[a[i]]=true;
			}
			
			int res[]=new int[2*n];
			boolean pos=true;
			for(int i=0;i<n;i++){
				int l=2*i;
				int r=2*i+1;
				/*if(vis[a[i]]){
					pos=false;
					break;
				}
				vis[a[i]]=true;*/
				res[l]=a[i];
				for(int j=a[i]+1;j<=2*n;j++){
					if(!vis[j]){
						vis[j]=true;
						res[r]=j;
						break;
					}
				}
				if(res[r]==0){
					pos=false;
					break;
				}
			}
			if(!pos)	
				System.out.println("-1");
			else{
				System.out.print(res[0]);
				for(int i=1;i<2*n;i++)
					System.out.print(" "+res[i]);
				System.out.println();
			}
		}
	}
}