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 Solution {
	
	public static void main(String[] args) {

		/*
		 * 
		 */
		Scanner scan = new Scanner(System.in);
		int t = scan.nextInt();
		while(t-->0) {
			int n = scan.nextInt();
			int[] arr = new int[(2*n)];
			boolean flag=false;
			int[] nums = new int[2*n+1];
			for(int i=0;i<2*n;i+=2) {
				int num=scan.nextInt();
				if(num>=2*n) flag=true;
				arr[i]=num;
				nums[num]++;
				if(nums[num]>1) flag=true;
			}
			if(flag) {
				System.out.println("-1");
				continue;
			}
			
			for(int i=0;i<2*n;i+=2) {
				int num = arr[i];
				for(int j=num+1;j<2*n+1;j++) {
					if(nums[j]==0) {
						nums[j]=1;
						arr[i+1]=j;
						break;
					}
				}
				if(arr[i+1]==0) {
					System.out.println(-1);
					flag=true;
					break;
				}
			}
			if(!flag) {
				for(int i=0;i<2*n;i++) {
					if(i>0) System.out.print(" ");
					System.out.print(arr[i]);
				}
				System.out.println();
			}
		}
		
		scan.close();
		
	}
	
}