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.Arrays;
import java.util.HashSet;
import java.util.Scanner;

public class restoringPermutation {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		StringBuilder finalAns = new StringBuilder();
		int t = sc.nextInt();
		while (--t >= 0) {
			StringBuilder ans = new StringBuilder();
			int n = sc.nextInt(), b[] = new int[n];
			HashSet<Integer> hs = new HashSet<Integer>(2 * n);
			for (int i = 0; i < n; i++) {
				b[i] = sc.nextInt();
				hs.add(b[i]);
			}
			int flag = 0;
			for (int i = 0; i < n; i++) {
				ans.append(b[i] + " ");
				int temp = b[i] + 1;
				while (hs.contains(temp)) {
					++temp;
				}
				if (temp > 2 * n) {
					flag = 1;
					break;
				}
				if (((2 * i) + 1) < 2 * n) {
					ans.append(temp + " ");
					hs.add(temp);
				}
			}
			finalAns.append(((flag == 0) ? ans : -1) + "\n");
		}
		sc.close();
		System.out.println(finalAns);
	}
}