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 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;



public class test {

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(br.readLine());
		StringTokenizer st;
		for(int i=0;i<t;i++) {
			int n=Integer.parseInt(br.readLine());
			st=new StringTokenizer(br.readLine());
			int[] arr=new int[2*n+1];
			boolean[] b=new boolean[2*n+1];
			boolean notpossible=false;
			for(int j=1;j<=n;j++) {
				arr[2*j-1]=Integer.parseInt(st.nextToken());
				b[arr[2*j-1]]=true;
			}
			if(!b[1]) {
				notpossible=true;
			}
			if(!notpossible) {
				for(int j=1;j<=n;j++) {
					int r=arr[2*j-1];
					r++;
					while(r<=2*n && b[r]) {
						r++;
					}
					if(r>2*n) {
						notpossible=true;
						break;
						
					}
					arr[2*j]=r;
					b[r]=true;
				}
			}
			if(!notpossible) {
				for(int j=1;j<=n;j++) {
					System.out.print(arr[2*j-1]+" "+arr[2*j]+" ");
				}
				System.out.println();
			}else {
				System.out.println(-1);
			}
		}
	}
	
}