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.BufferedReader;
import java.io.InputStreamReader;

public class Solution {
    public static void main(String[] args){
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            int T = Integer.parseInt(br.readLine());
            while (T != 0) {
                T--;
                int n= Integer.parseInt(br.readLine());
                int [] b = new int[n+1];
                boolean [] vis = new boolean[2*(n+1)];
                String [] si = br.readLine().split(" ");
                for(int i=0;i<n;i++){
                    vis[2*i] = false;
                    vis[2*i +1] = false;
                }
                for(int i=0;i<n;i++){
                    b[i] = Integer.parseInt(si[i]);
                    vis[b[i]-1]=true;
                }
                boolean impossible = false;
                String res = "";
                for(int i=0;i<n;i++){
                    if(i!=0)
                        res +=" ";
                    res+=b[i];
                    String tmp = "";
                    for(int j = b[i];j<2*n;j++){
                        if(!vis[j]){
                            vis[j]=true;
                            tmp=" "+Integer.toString(j+1);
                            break;
                        }
                    }
                    if(tmp == ""){
                        impossible = true;
                        break;
                    }else{
                        res+=tmp;
                    }
                }
                if(impossible){
                    System.out.println(-1);
                }else{
                    System.out.println(res);
                }
            }
        }catch (Exception e){
            
        }
    }
}