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.util.*;
public class Solutions {
	static Scanner scr=new Scanner(System.in);
	public static void main(String[] args) {
		int t=read();
		while(t-->0) {
			solve();
		}
	}
	static void solve() {
		int n=read();
		int []a=new int[n];
		int []count=new int[2*n+1];
		for(int i=0;i<n;i++) {
			a[i]=read();
			count[a[i]]=1;
		}
		int []res=new int[2*n];
		for(int i=0;i<n;i++) {
			res[2*i]=a[i];
		}
		int index=1;
		for(int i=0;i<n;i++) {
			int num=a[i];
			int curr=a[i]+1;
			while(curr<=2*n) {
				if(count[curr]==0) {
					count[curr]=1;
					res[index]=curr;
					index+=2;
					break;
				}else {
					curr++;
				}
			}
			if(curr==2*n+1) {
				println(-1);
				return;
			}
		}
		printArray(res);
		println();
	}
	static int[] sort(int a[]) {
		Arrays.sort(a);
		return a;
	}
	static String reverse(String s) {
		String res="";
		for(int i=s.length()-1;i>=0;i--) {
			res+=s.charAt(i);
		}
		return res;
	}
	static int gcd(int a,int b) {
		if(b==0) {
			return a;
		}
		return gcd(b,a%b);
	}
	static void swap(int a[],int i,int j) {
		int temp=a[i];
		a[i]=a[j];
		a[j]=temp;
	}
	static boolean contains(HashMap<Integer,Integer>hm,int ele) {
		return hm.containsKey(ele);
	}static boolean contains(HashMap<Character,Integer>hm,char ele) {
		return hm.containsKey(ele);
	}static int get(HashMap<Character,Integer>hm,char ele) {
		return hm.get(ele);
	}
	static char[] cha(String s) {
		return s.toCharArray();
	}
	static int len(String s) {
		return s.length();
	}
	static int len(int a[]) {
		return a.length;
	}
	static int len(long a[]) {
		return a.length;
	}
	static long abs(long a) {
		return Math.abs(a);
	}static int abs(int a) {
		return Math.abs(a);
	}
	static long min(long a,long b) {
		return Math.min(a, b);
	}static long max(long a,long b) {
		return Math.max(a, b);
	}
	static void print(int a) {
		System.out.print(a+" ");
	}static void print(long a) {
		System.out.print(a+" ");
	}static void print(String a) {
		System.out.print(a);
	}static void println(int a) {
		System.out.println(a);
	}static void println(long a) {
		System.out.println(a);
	}static void println(String a) {
		System.out.println(a);
	}
	static void println() {
		System.out.println();
	}
	static void println(char []c) {
		for(int i=0;i<c.length;i++) {
			System.out.print(c[i]);
		}
		println();
	}
	static int read() {
		return scr.nextInt();
	}
	static String readS() {
		return scr.next();
	}
	static long readL() {
		return scr.nextLong();
	}
	static int[]readIntArray(int n){
//		int n=read();
		int []a=new int[n];
		for(int i=0;i<n;i++) {
			a[i]=scr.nextInt();
		}
		return a;
	}static long[]readLongArray(int n){
		long []a=new long[n];
		for(int i=0;i<n;i++) {
			a[i]=scr.nextLong();
		}
		return a;
	}
	static void printArray(int a[]) {
		int n=a.length;
		for(int i=0;i<n;i++) {
			print(a[i]);
		}
	}
}
