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;

public class C1315 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int ntc = sc.nextInt();
        while(ntc-->0){
            int n = sc.nextInt();
            int[] arr = new int[n];
            int twiceN = 2*n +1;
            int[] arrTwice = new int[twiceN];
            arrTwice[0] = -1;
            StringBuilder ansString = new StringBuilder();
            boolean hasAns = true;
            for(int i = 0;i<n;i++){
                arr[i] = sc.nextInt();
            }
            for(int i=0;i<n;i++){
                arrTwice[arr[i]] = -1;
            }
            int j;
            for(int i=0;i<n;i++){
                ansString.append(arr[i]+" ");
                for(j=arr[i]+1;j<twiceN;j++){
                    if(arrTwice[j]!=-1){
                        ansString.append(j+" ");
                        arrTwice[j]=-1;
                        break;
                    }
                }
                if(j == twiceN){
                    hasAns = false;
                    break;
                }
            }
            if(!hasAns){
                System.out.println(-1);
            }else{
                System.out.println(ansString);
            }
        }
    }
}