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.HashSet;
import java.util.Scanner;

public class B16 {
	static int max = 11;

	public static void main(String[] args) {

		Scanner nik = new Scanner(System.in);
		int t = nik.nextInt();
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < t; i++) {

		}
		A: while (t-- > 0) {
			int n = nik.nextInt();
			int b[] = new int[n + 1];
			HashSet<Integer> hs = new HashSet<>();
			for (int i = 1; i <= n; i++) {
				b[i] = nik.nextInt();
				hs.add(b[i]);
			}
			int a[] = new int[2 * n + 1];
			int val = 1;
			for (int i = 1; i <= n; i++) {
				int v = b[i];
				a[2 * i - 1] = v;
				val = v + 1;
				while (hs.contains(val))
					val++;
				if (val > 2 * n) {
					sb.append(-1 + "\n");
					continue A;
				}
				hs.add(val);
				a[2 * i] = val;
			}
			for (int i = 1; i < a.length; i++)
				sb.append(a[i] + " ");
			sb.append("\n");
		}
		System.out.print(sb);
	}

}