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.util.Scanner;
import java.util.Stack;
import java.util.Arrays;
public class C{
	public static void main(String[] args){
			Scanner in = new Scanner(System.in);
			int t = in.nextInt();
			StringBuilder res = new StringBuilder();
			o1: for(int j = 0;j<t;++j){
				StringBuilder resLocal  = new StringBuilder();
				int n = in.nextInt();
				int[] b = new int[n];
				for(int i = 0;i<n;++i) b[i] = in.nextInt();
				int[] a = new int[2*n];
				int[] flag = new int[2*n+1];
				for(int i = 0;i<n;++i){
					if(flag[b[i]] == 1){
						res.append("-1\n");
						continue o1;
					}
					a[2*i] = b[i];
					flag[b[i]] = 1;
				}
				// System.out.println(Arrays.toString(a) + "\n");
				for(int i = 0;i<2*n;++i){
					if(a[i] == 0) {
						for(int k = a[i-1];k<=2*n;++k){
							if(flag[k] == 1) continue;
							// System.out.println(flag[k] + " " + k +  " " + i);
							flag[k] = 1;
							a[i] = k;
							break;
						}
					}
					if(a[i] == 0){
						res.append("-1\n");
						continue o1;
					}
					resLocal.append(a[i] + " ");
				}
				res.append(resLocal.append("\n").toString());
			}
			System.out.println(res);
						
	}
}