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.Scanner;

public class C {

	public static void main(String[] args) {
		new C();
	}

	public C() {
		Scanner sc = new Scanner(System.in);

		int amountT = sc.nextInt();
		for (int task = 0; task < amountT; task++) {
			int n = sc.nextInt();
			int[] b = new int[n];
			boolean[] done = new boolean[2 * n + 1];
			boolean poss = true;
			for (int i = 0; i < n; i++) {
				int val = sc.nextInt();
				b[i] = val;
				if (done[val]) {
					poss = false;
				}
				done[val] = true;
			}
			int[] a = new int[n];
			for (int index = 0; index < n; index++) {
				boolean found = false;
				for (int i = b[index] + 1; i <= 2*n; i++) {			
					if (!done[i]) {
						if (i > b[index]) {
							found = true;
							a[index] = i;
							done[i] = true;
							break;
						}
					}
				}
				if (!found) {
					poss = false;
					break;
				}
			}
			if (poss) {
				String s = "";
				for (int i = 0; i < n; i++) {
					if (i != 0) {
						s += " ";
					}
					s += b[i] + " " + a[i];
				}
				System.out.println(s);
			} else {
				System.out.println(-1);
			}
		}

		sc.close();
	}
}