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 test {
    static int m[];
    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        int t=sc.nextInt();
        while (t-->0){
            StringBuilder sb=new StringBuilder();
            int N=sc.nextInt();
            int sum[]=new int[N];
            boolean con[]=new boolean[2*N + 1];
            TreeSet<Integer> set=new TreeSet<>();
            for (int i=0;i<N;i++){
                sum[i]=sc.nextInt();
                con[sum[i]]=true;
            }
            for (int i=1;i<=2*N;i++){
                if (!con[i])set.add(i);
            }
            boolean z=false;
            for (int j=0;j<N;j++) {
                Integer x=set.ceiling(sum[j]);
                if (x==null){
                    z=true;
                    break;
                }
                sb.append(sum[j]+" "+x+" ");
                set.remove(x);
            }
            if (z) System.out.println(-1);
            else System.out.println(sb);
        }
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}