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.*;

//odd selsection 1363-A

public class CodeForce {
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int[] b = new int[n];
            for (int i = 0; i < b.length; i++) {
                b[i] = sc.nextInt();
            }
            int[] arr = ultimate(b, n);
            if (arr == null)
                System.out.println(-1);
            else
                printarr(arr);
        }
    }

    public static void printarr(int[] a) {
        for (int i = 0; i < a.length; i++) {
            System.out.print(a[i] + " ");
        }
        System.out.println();
    }

    public static int[] ultimate(int[] b, int n) {
        int[] a = new int[2 * n];
        boolean[] vis = new boolean[2 * n + 1];
        for (int i = 0; i < b.length; i++) {
            a[(i + 1) * 2 - 2] = b[i];
            vis[b[i]] = true;
        }
        for (int i = 1; i < a.length; i += 2) {
            for (int j = 1; j < vis.length; j++) {
                if (!vis[j] && j > a[i - 1]) {
                    a[i] = j;
                    vis[j] = true;
                    break;
                }
            }
        }
        for (int i = 1; i < vis.length; i++) {
            if (!vis[i])
                return null;
        }
        return a;
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            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;
        }
    }
}