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 RestoringPermutation {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		while(T-->0){
			int n = sc.nextInt();
			int arr[] = new int[n];
			int covered[] = new int[2*n+1];
			for(int i=0;i<n;i++){
				int num = sc.nextInt();
				arr[i] = num;
				covered[num] = 1;
			}
			int ans[] = new int[n];
			int count = 0;
			for(int i=0;i<n;i++){
				int num = arr[i];
				for(int j=num+1;j<=2*n;j++){
					if(covered[j] == 0){
						ans[i] = j;
						covered[j] = 1;
						count++;
						break;
					}
				}
			}
			if(count == n){
				for(int i=0;i<n;i++){
					System.out.print(arr[i]+" "+ans[i]+" ");
				}
				System.out.println();
			}
			else
				System.out.println("-1");
		}
		
	}

}