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.*;
import java.lang.*;

public class c1 {
    public static FScanner scan;
    public static PrintWriter out;
    public static void main(String[] args) {
        scan=new FScanner();
        out=new PrintWriter(new BufferedOutputStream(System.out));

//        int t=1;
        int t=scan.nextInt();
        while(t-->0) {

            int n=scan.nextInt();
            int[] a=new int[n],ans=new int[2*n];
            ArrayList<Integer> unused=new ArrayList<>();
            for(int c=0;c<2*n;c++) unused.add(c+1);
            for(int c=0;c<n;c++) {
                a[c]=scan.nextInt();
                ans[2*c]=a[c];
                unused.remove(unused.indexOf(a[c]));
            }

            boolean possible=true;
            for(int c=0;c<n;c++) {
                int num=bs(ans[2*c],unused);
                if(num==-1) {
                    possible=false;
                    break;
                } else {
                    ans[2*c+1]=unused.get(num);
                    unused.remove(num);
                }
            }

            if(!possible) out.println(-1);
            else {
                for(int c=0;c<2*n;c++) out.print(ans[c]+" ");
                out.println();
            }
        }
        out.close();
    }

    //-----------------------------------------------------HERE---------------------------------------------------------

    static int bs(int min,ArrayList<Integer> list) {
        int start=0,end=list.size()-1,ans=-1;
        while(start<=end) {
            int mid=(start+end)/2;
            if(list.get(mid)>=min) {
                ans=mid;
                end=mid-1;
            } else start=mid+1;
        } return ans;
    }


    //------------------------------------------------------------------------------------------------------------------

    //scanner
    public static class FScanner {
        BufferedReader br;
        StringTokenizer st;

        public FScanner() {
            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;
        }
    }
}