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 static java.lang.System.*;

public class Main
{
    public static void main(String[] args) {
        Scanner s = new Scanner(in);
        int t = s.nextInt();
        s.nextLine();

        while (t-- > 0) {
            int n = s.nextInt();
            int[] a = new int[n];
            boolean[] taken = new boolean[n * 2 + 1];
            boolean tooLarge = false;
            int largest = 0;
            for (int i = 0; i < n; i++) {
                int temp = s.nextInt();
                if (temp >= n * 2) {
                    tooLarge = true;
                    break;
                }
                a[i] = temp;
                taken[temp] = true;
                if (temp > largest) largest = temp;
            }
            s.nextLine();
            if (tooLarge) {
                out.println(-1);
                continue;
            }

            boolean fail = false;
            int usableNums = 0;
            for (int i = 1; i < largest; i++) {
                if (!taken[i]) usableNums--;
                else usableNums++;

                if (usableNums < 0) {
                    fail = true;
                    break;
                }
            }
            if (fail) {
                out.println(-1);
                continue;
            }

            for (int i = 0; i < n; i++) {
                out.print(a[i] + " ");
                int temp = a[i] + 1;
                while (taken[temp]) temp++;
                taken[temp] = true;
                out.print(temp + " ");
            }
            out.println();
        }
    }
}