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 C1315 {
	

	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int T=fs.nextInt();
		for (int tt=1; tt<=T; tt++) {
			int n=fs.nextInt();
			int firstNum[]=fs.readArray(n);
			int secondNum[]=new int [n];
			int counter[]=new int [2*n+1];
			for(int i=0;i<n;i++) {
				counter[firstNum[i]]++;
			}
			boolean isPossible=true;
			for(int i=0;i<n;i++) {
				for(int j=firstNum[i];j<=2*n;j++) {
					if(counter[j]==0) {
						counter[j]++;
						secondNum[i]=j;
						break;
					}
				}
			}
			for(int i=1;i<=2*n;i++) {
				if(counter[i]==0) {
					isPossible=false;
					break;
				}
			}
			if(isPossible) {
				for(int i=0;i<n;i++) {
					System.out.print(firstNum[i]+" "+secondNum[i]+" ");
				}
				System.out.println();
			}
			else {
				System.out.println(-1);
			}
		}
	}

	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();
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			}
			catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		int[][] readArray(int n,int m) {
			int[][] a=new int[n][m];
			for (int i=0; i<n; i++) for(int j=0; j<m; j++) a[i][j]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

	
}