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.*;

public class RestoringPermutation {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0){
			int n = sc.nextInt();
			int b[] = new int[n];
			int a[] = new int[n*2];
			int r[] = new int[n*2];
			boolean bool = false;
			for(int i=0;i<n;i++){
				a[2*i] = 2*i + 1;
				a[2*i+1] = 2*i + 2;
				b[i] = sc.nextInt();
			}
			for(int i=0;i<n;i++){
				a[b[i] - 1] = 0;
			}
			for(int i=0;i<n;i++){
				r[2*i] = b[i];
				bool = false;
				for(int j=0;j<2*n;j++){
					if(a[j] > 0 && a[j] > b[i]){
						r[2*i+1] = a[j];
						a[j] = 0;
						bool = true;
						break;
					}
				}
				if(!bool){
					System.out.println(-1);
					break;
				}
			}
			if(bool){
				for(int i=0;i<2*n;i++){
					System.out.print(r[i] + " ");
				}
				System.out.println();
			}
		}
	}

}