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.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;

public class C implements Runnable {
		
	
	private void solve() throws IOException {
		int t = nextInt();
		outer2: while(t-- > 0) {
			int n = nextInt();
			int[] b = new int[n];
			for (int i = 0; i < n; i++) {
				b[i] = nextInt();
			}
			boolean[] vis = new boolean[2 * n + 1];
			int[] ans = new int[2 * n];
			for (int i = 0; i < b.length; i++) {
				ans[2 * i] = b[i];
				vis[b[i]] = true;
			}
			
			outer: for (int i = 1; i < ans.length; i += 2) {
				for (int j = ans[i - 1] + 1; j <= 2 * n; j++) {
					if (!vis[j]) { 
						ans[i] = j;
						vis[j] = true;
						continue outer;
					}
				}
				pl(-1); continue outer2;
			}
			
			for (int o : ans) p(o + " ");
			pl();
		}
	}

	public static void main(String[] args) {
		new C().run();
	}

	BufferedReader reader;
	StringTokenizer tokenizer;
	PrintWriter writer;

	public void run() {
		try {
			reader = new BufferedReader(new BufferedReader(new InputStreamReader(System.in)));
			writer = new PrintWriter(System.out);
			tokenizer = null;
			solve();
			reader.close();
			writer.close();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	BigInteger nextBigInteger() throws IOException {
		return new BigInteger(nextToken());
	}

	String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}

	void p(Object... objects) {
		for (int i = 0; i < objects.length; i++) {
			if (i != 0)
				writer.print(' ');
			writer.print(objects[i]);
		}
	}

	void pl(Object... objects) {
		p(objects);
		writer.println();
	}

	int cc;

	void pf() {
		writer.printf("Case #%d: ", ++cc);
	}

}