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 MyClass {
    public static void main(String args[]) {
      Scanner sc = new Scanner(System.in);
      int t = sc.nextInt();
      while(t-->0)
      {
          int n = sc.nextInt();
          int[] a = new int[n];
          int[] b = new int[2*n+1];
          for(int i=0;i<n;i++)
          {
              a[i] = sc.nextInt();
              b[a[i]]=1;
          }
          
          ArrayList<Integer> arr = new ArrayList<Integer>();
         
          for(int i=0;i<n;i++)
          {
              arr.add(a[i]);
              int d = 0;
              for(int j=1;j<2*n+1;j++)
              {
                 
                  if(j>a[i]&&b[j]==0)
                  {
                      d++;
                      arr.add(j);
                      b[j]=1;
                      break;
                  }
              }
              if(d==0)
              {
                  arr.add(0);
              }
          }
          
          int dd=0;
          for(int i=0;i<2*n;i++)
          {
              if(arr.get(i)==0)
              {
                  dd++;
                  break;
              }
          }
          if(dd==1)
          {
              System.out.println(-1);
          }
          else
          {
              for(int i=0;i<2*n;i++)
              {
                  System.out.print(arr.get(i)+" ");
              }
              System.out.println();
          }
          
    
      }
    }
}