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.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;
import java.text.*;
import java.math.*;
import java.util.*;

public class Main implements Runnable {
	
	public int getMin(boolean isUsed[], int val, int n){
		
		int min=-1;
		for(int v=val;v<=2*n;v++){
			if(isUsed[v]==false){
				min=v;
				isUsed[v]=true;
				break;
			}
		}
		return min;
	}
	
	public void solve() throws Exception {

		int t=iread();
		for(int cases=1;cases<=t;cases++){
			int n=iread();
			int []b= new int[n+1];
			boolean isUsed[]= new boolean[2*n+1];
			Arrays.fill(isUsed, false);
			boolean hasResult=true;
			for(int i=1;i<=n;i++){
				b[i]= iread();
				if(b[i]>2*n || isUsed[b[i]])
					hasResult=false;
				else
					isUsed[b[i]]=true;
			}
			if(!hasResult){
				System.out.println("-1");
				continue;
			}
			
			int []result= new int[2*n+1];
			Arrays.fill(isUsed, false);
			
			for(int i=1;i<=n;i++){
				result[2*i-1]=b[i];
				isUsed[b[i]]=true;
			}
			for(int i=2;i<=2*n;i+=2){
				
				int x= getMin(isUsed, result[i-1], n);
				if(x==-1){
					hasResult=false;
					break;
				}
				else
					result[i]=x;
			}
			
			if(!hasResult){
				System.out.println("-1");
			}
			else{
				for(int i=1;i<=2*n;i++)
					System.out.print(result[i]+" ");
			}
			
		}

		
		
		
	}


	

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new BufferedWriter(new OutputStreamWriter(System.out));
			// in = new BufferedReader(new FileReader(filename+".in"));
			// out = new BufferedWriter(new FileWriter(filename+".out"));
			solve();
			out.flush();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public int iread() throws Exception {
		return Integer.parseInt(readword());
	}

	public double dread() throws Exception {
		return Double.parseDouble(readword());
	}

	public long lread() throws Exception {
		return Long.parseLong(readword());
	}

	BufferedReader in;

	BufferedWriter out;

	public String readword() throws IOException {
		StringBuilder b = new StringBuilder();
		int c;
		c = in.read();
		while (c >= 0 && c <= ' ')
			c = in.read();
		if (c < 0)
			return "";
		while (c > ' ') {
			b.append((char) c);
			c = in.read();
		}
		return b.toString();
	}

	public static void main(String[] args) {

		try {
			Locale.setDefault(Locale.US);
		} catch (Exception e) {

		}
		// new Thread(new Main()).start();
		new Thread(null, new Main(), "1", 1 << 25).start();
	}
}