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.HashMap;
import java.util.Scanner;

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while (T-- > 0) {
            int a = sc.nextInt();
            int[] x = new int[a + 1];
            int min = 1;
            int[] y = new int[a * 2 + 1];
            boolean bo = true;
            HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
            for (int i = 1; i <= a; i++) {
                x[i] = sc.nextInt();
                map.put(x[i], 1);
            }
            for (int i = 1; i <= a; i++) {
                int k = -1;
                for (int j = x[i] + 1; j <= 2 * a; j++) {
                    if (!map.containsKey(j)) {
                        map.put(j, 1);
                        k = j;
                        break;
                    }
                }
                if (k == -1) {
                    bo = false;
                    System.out.println(-1);
                    break;
                } else {
                    y[i * 2] = k;
                    y[i * 2 - 1] = x[i];
                }
            }
            if (bo) {
                for (int i = 1; i < 2 * a; i++) {
                    System.out.print(y[i] + " ");
                }
                System.out.println(y[2 * a]);
            }

        }
    }
}