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 
import java.io.*;
import java.util.*;

public class Permutation {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        for (int i1 = 0; i1 < t; i1 ++){
            int n = sc.nextInt();
            int[] a = new int[2 * n];
            boolean[] taken = new boolean[2 * n + 1];
            for (int i2 = 0; i2 < n; i2 ++){
                a[i2 * 2] = sc.nextInt();
                taken[a[i2 * 2]] = true;
            }
            for (int i3 = 0; i3 < n; i3 ++){
                for (int i4 = a[i3 * 2] + 1; i4 <= 2 * n; i4 ++){
                    if (!taken[i4]){
                        a[i3 * 2 + 1] = i4;
                        taken[i4] = true;
                        break;
                    }
                }
            }
            boolean works = true;
            StringBuilder ans = new StringBuilder();
            for (int i5 = 0; i5 < 2 * n; i5 ++){
                ans.append(a[i5]);
                ans.append(" ");
                if (a[i5] == 0){
                    pw.println(-1);
                    works = false;
                    break;
                }
            }
            if (works){
                pw.println(ans);
            }
        }
        pw.close();
    }
}
