Problem: Permutation p is a sequence of integers p=[p_1, p_2, ..., p_n], consisting of n distinct (unique) positive integers between 1 and n, inclusive. For example, the following sequences are permutations: [3, 4, 1, 2], [1], [1, 2]. The following sequences are not permutations: [0], [1, 2, 1], [2, 3], [0, 1, 2].

The important key is in the locked box that you need to open. To open the box you need to enter secret code. Secret code is a permutation p of length n. 

You don't know this permutation, you only know the array q of prefix maximums of this permutation. Formally:

  * q_1=p_1, 
  * q_2=max(p_1, p_2), 
  * q_3=max(p_1, p_2,p_3), 
  * ... 
  * q_n=max(p_1, p_2,...,p_n). 



You want to construct any possible suitable permutation (i.e. any such permutation, that calculated q for this permutation is equal to the given array).

Input

The first line contains integer number t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.

The first line of a test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of elements in the secret code permutation p.

The second line of a test case contains n integers q_1, q_2, ..., q_n (1 ≤ q_i ≤ n) — elements of the array q for secret permutation. It is guaranteed that q_i ≤ q_{i+1} for all i (1 ≤ i < n).

The sum of all values n over all the test cases in the input doesn't exceed 10^5.

Output

For each test case, print:

  * If it's impossible to find such a permutation p, print "-1" (without quotes). 
  * Otherwise, print n distinct integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). If there are multiple possible answers, you can print any of them. 

Example

Input


4
5
1 3 4 5 5
4
1 1 3 4
2
2 2
1
1


Output


1 3 4 5 2 
-1
2 1 
1 

Note

In the first test case of the example answer [1,3,4,5,2] is the only possible answer:

  * q_{1} = p_{1} = 1; 
  * q_{2} = max(p_{1}, p_{2}) = 3; 
  * q_{3} = max(p_{1}, p_{2}, p_{3}) = 4; 
  * q_{4} = max(p_{1}, p_{2}, p_{3}, p_{4}) = 5; 
  * q_{5} = max(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}) = 5. 



It can be proved that there are no answers for the second test case of the example.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main{ 
   static class Fast{
     public BufferedReader br;
     public StringTokenizer st;
     
     public Fast(){
          br =new BufferedReader(new InputStreamReader(System.in));
     }
     
     String next(){
          while(st==null || !st.hasMoreTokens()){
               try{
                    st=new StringTokenizer(br.readLine());
               }
               catch(IOException e){
                    throw new RuntimeException(e);
               }
               
          }
               return st.nextToken();
          }
     int nextInt(){
          return Integer.parseInt(next());
     }

     long nextLong(){
          return Long.parseLong(next());
     }

 }
  
 

   static void sort(int[] a) {
      ArrayList<Integer> list=new ArrayList<>();
      for (int i:a) list.add(i);
      Collections.sort(list);
      for (int i=0; i<a.length; i++) a[i]=list.get(i);
    }   
	public static void main(String[] args)throws Exception{
         //StringTokenizer st;//! @ % & * () _ {} # ~ : < > ? "" | ^
          Fast f=new Fast();
         PrintWriter out = new PrintWriter(System.out);
      
         //  int x=Integer.parseInt(st.nextToken());
         
        //st=new StringTokenizer(br.readLine()); 
         int t=f.nextInt();
         while(t-->0){

           int n=f.nextInt();
           int q[]=new int[n],a[]=new int[n];

           boolean h[]=new boolean[n+10];
           boolean k=false;
           //int h[]=new int[5];
           for (int i=0;i<n;i++) {
             q[i]=f.nextInt();
           }
           int prevj=1;
           a[0]=q[0];
           h[a[0]]=true;
           for (int i=1;i<n;) {
             if (!h[q[i]]) {
                a[i]=q[i];
                h[a[i]]=true;
                i++;
             }
             else{
                       for (int j=prevj;j<=q[i];j++) {
                         if (!h[j]) {a[i]=j;h[a[i]]=true;prevj=j+1;break;}
                       }
                       if (a[i]==0) {k=true;break;}
                         if (k) {
                           break;
                         }
                       i++;
             }
           }

           if (k) {
             out.println(-1);
           }

           else {

            for (int i=0;i<n;i++) {
              out.print(a[i]+" ");
            }
            out.println();
           }
            
           

           
           //out.println(ans);


        }
        //out.println(mh);
    out.close();

}
}