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.math.*;
import java.util.*;
import java.lang.*;
public class RestoringPermutation { 
 
	public static void main(String[] args) { 
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		int T = I.nextInt();
		while (T-->0) {
			int N = I.nextInt();
			boolean[] vis = new boolean[2*N+1];
			int[] ans = new int[2*N];
			int[] a = new int[N];
			boolean bad = false;
			for (int i = 0; i < N; i++) {
				a[i] = I.nextInt();
				vis[a[i]] = true;
			}
			for (int cur = 0; cur < N; cur++) {
				int conv = cur+1; //Conversion to one-based indexing
				int L = conv*2-1;
				int R = conv*2;
				L--;
				R--; //Converting back to 0 based indexing
				ans[L] = a[cur]; //The minimum should come first if lexicographically minimal
				//solution is needed
				boolean added=false;
				for (int pair = a[cur]+1; pair <= 2*N; pair++) { //Looking for larger partner
					if (!vis[pair]) {
						added=true;
						ans[R]=pair;
						vis[pair]=true;
						break;
					}
				}
				if (!added) {
					bad=true;
					break;
				}
			}
			if (bad) O.pln(-1);
			else {
				for (int i = 0; i < 2*N; i++) O.p(ans[i]+" ");
				O.p("\n");
			}
		}
	}
	public static void sort(long[] a) {Arrays.sort(a);}
	public static void sort(int[] a) {Arrays.sort(a);}
	public static void sort(String[] a) {Arrays.sort(a);}
	public static void sort(char[] a) {Arrays.sort(a);}
	public static long ceil(long num, long den) {long ans = num/den; if (num%den!=0) 
	ans++; return ans;}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return Math.max(a,b);
		return GCD(Math.min(a, b),Math.max(a, b)%Math.min(a, b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());};
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void pln(StringBuilder x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
	}
}