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.*;

public class code_forces {

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

		Scanner s= new Scanner(System.in);	

		int t= s.nextInt();

		outer: while(t-->0) {

			int n= s.nextInt();
			int[] arr= new int[n];

			for(int i=0; i<arr.length;i++) {
				arr[i] = s.nextInt();
			}

			int[] help= new int[2*n+1];
			int[] ans = new int[2*n];

			for(int i=0; i<arr.length;i++) {
				ans[2*i]= arr[i];
				help[arr[i]]=1;
			}

			for(int i=0; i<ans.length;i++) {

				if(ans[i]==0) {

					int j= ans[i-1];
					while(j<help.length && help[j]!=0) {
						j++;
					}

					if(j<help.length && help[j]==0) {
						ans[i] = j;
						help[j]=1;
					}
					else {
						System.out.println("-1");
						continue outer;
					}
				}
			}

			for(int i=0; i<ans.length;i++) {
				System.out.print(ans[i]+" ");
			}
			System.out.println();
		}
	}
}