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 Perms {

	public static void main(String[] args) {
		Scanner scanner= new Scanner(System.in);
		int tCount= scanner.nextInt();
		int [][]cases= new int[tCount][];
		for(int i=0; i<tCount;i++){
			int itemsCount= scanner.nextInt();
			int[]arr= new int[itemsCount];
			for(int j=0;j<itemsCount;j++){
				arr[j]=scanner.nextInt();
			}
			cases[i]=arr;
		}
		for (int i=0; i<cases.length;i++){
			printArr(perms(cases[i]));
		}
	}
	
	static void printArr(int[]arr){
		for (int i=0; i<arr.length;i++){
			System.out.print(arr[i]+" ");
		}
		System.out.println();
	}
	
	static int[] perms(int []a){
		int len= a.length;
		int[] noResult={-1};
		int[] result= new int [2*len];
		if(max(a,len)>2*len||find(a,len,2*len)){
			return noResult;
		}
		for(int i=0;i<len;i++){
			result[2*i]=a[i];
		}
		for(int i=0;i<len;i++){
			int ind=2*i+1;
			for(int j=1;j<=2*len;j++){
				if((!find(result,2*len,j))&&j>result[ind-1]){
					result[ind]=j;
					break;
				}
			}
			if(result[ind]==0){
				return noResult;
			}
		}
		return result;
	}
	
	static int max(int[]arr, int len){
		int m= arr[0];
		for(int i=1;i<len;i++){
			if(arr[i]>m){
				m= arr[i];
			}
		}
		return m;
	}
		
	static boolean find(int[]arr, int len, int x){
		for(int i=0;i<len;i++){
			if(arr[i]==x){
				return true;
			}
		}
		return false;
	}
}