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.io.BufferedReader;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Stack;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int a[]=new int[210];
			for(int i=1;i<=200;i++) a[i]=0;
			int b[]=new int[101];
			int ans[]=new int[210];
			int n=sc.nextInt();
			for(int i=1;i<=n;i++) {b[i]=sc.nextInt();a[b[i]]=1;}
			for(int i=1;i<=n;i++) {
				for(int j=b[i]+1;j<=2*n;j++) {
					if(a[j]==0) {ans[2*i-1]=b[i];ans[2*i]=j;a[j]=1;break;} 
				}
			}
			int g=1;
			for(int i=1;i<=2*n;i++) if(a[i]==0) g=0;
			if(g==0) System.out.println(-1);
			else 
				{for(int i=1;i<=2*n;i++) System.out.print(ans[i]+" ");
				System.out.println();
				}
		}
		sc.close();
	}
}