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:
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1315C extends PrintWriter {
	CF1315C() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1315C o = new CF1315C(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] bb = new int[n];
			boolean[] used = new boolean[n * 2 + 1];
			for (int i = 0; i < n; i++) {
				int b = sc.nextInt();
				bb[i] = b;
				used[b] = true;
			}
			int[] aa = new int[n * 2];
			boolean yes = true;
			for (int i = 0; i < n; i++) {
				int b = bb[i];
				int c = b + 1;
				while (c <= n * 2 && used[c])
					c++;
				if (c > n * 2) {
					yes = false;
					break;
				}
				used[c] = true;
				aa[i << 1] = b;
				aa[i << 1 | 1] = c;
			}
			if (yes) {
				for (int i = 0; i < n * 2; i++)
					print(aa[i] + " ");
				println();
			} else
				println(-1);
		}
	}
}