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.*;

public class Main {

    /**
     *
     * k = 1
     *
     * 2^17+1  2^17     0
     * 1      2^17+1   1
     *
     * dp[2][2] = max(
     */

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        int t = sc.nextInt();
        outer: while (t-->0) {
          int n = sc.nextInt();
          int[] arr = new int[n + 1];
          for (int i = 1; i <= n; i++) {
              arr[i] = sc.nextInt();
          }
          int[] ans = new int[2 * n + 1];
            boolean[] used = new boolean[2*n+1];
            for (int i = 1; i <= n; i++) {
                ans[2 * i - 1] = arr[i];
                used[arr[i]] = true;
            }
            for (int i = 2; i <= 2 * n; i+=2) {
                int val = ans[i-1] + 1;
                while (val <= 2 * n && used[val]) {
                    val++;
                }
                if (val > 2 * n) {
                    System.out.println(-1);
                    continue outer;
                }
                used[val] = true;
                ans[i] = val;
            }

            for (int i = 1; i <= 2*n; i++) {
                System.out.print(ans[i] + " ");
            }
            System.out.println();
        }
    }

//    private static void solve(int[] ans, int n, int[] arr) {
//        boolean[] used = new boolean[2*n+1];
//        for (int i = 1; i <= n; i++) {
//            ans[2 * i - 1] = arr[i];
//            used[arr[i]] = true;
//        }
//        for (int i = 2; i <= 2 * n; i+=2) {
//            int val = ans[i-1] + 1;
//            while (val <= 2 * n && used[val]) {
//                val++;
//            }
//            if (val > 2 * n) {
//                System.out.println(-1);
//                return;
//            }
//            used[val] = true;
//            ans[i] = val;
//        }
//
//        for (int i = 1; i <= 2*n; i++) {
//            System.out.print(ans[i] + " ");
//        }
//        System.out.println();
//    }

    /**\
     *
     * 4
     * 2 3 4 5
     *
     *
     * 1 2 3 4 5 6 7 8
     * 2 6 3 7 4 8 5 9
     */

    // 1 1 1 0 0 0
    // 0 0 1 1 1 1


    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;
        }


    }
}

/**
 * 7 3
 * 3 1 5
 * 9 6 5
 *
 * i = 1: 1
 * i = 2: 2
 * i = 3: 3
 *
 *
 * 1 2 3 3 3 3 3
 */


/**
 *
 *  1+1
 **/