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.awt.Desktop;


import java.io.BufferedReader;


import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.URI;
import java.net.URISyntaxException;
import java.sql.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.LinkedHashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.util.Vector;



public  class codechef3 {
static class comp implements Comparator<Integer>
{

	@Override
	public int compare(Integer o1, Integer o2) {
		if(Math.abs(o1)>Math.abs(o2))
			return -1;
		else return 1;
	
	}
	
	}
      //=======================================================
	  //sorting Pair
   static class comp1 implements Comparator<Pair<Integer,Integer>>
   {

	@Override
	public int compare(Pair<Integer, Integer> o1, Pair<Integer, Integer> o2) {
		if(o1.k>o2.k)
			return 1;
		else return -1;
	
	}

	
	
	}
        //======================================================= 
	   
   //Creating Pair class  
         //----------------------------------------------------------------------
	    
       static class Pair<Integer,Intetger>
       {
    	   int k=0;
    	   int v=0;
    	   public Pair(int a,int b)
    	   {
    		   k=a;
    		   v=b;
    	   }
    	   
    	   
    	   
       }
    
     //--------------------------------------------------------------------------
	
		 static class FastReader
		    {BufferedReader br; 
		        StringTokenizer st; 
		 public FastReader() 
		        {  br = new BufferedReader(new
		                     InputStreamReader(System.in)); 
		        } 
		  
		       String next() 
		        { 
		            while (st == null || !st.hasMoreElements()) 
		            { 
		                try
		                { 
		                    st = new StringTokenizer(br.readLine()); 
		                } 
		                catch (IOException  e) 
		                { 
		                    e.printStackTrace(); 
		                } 
		            } 
		            return st.nextToken(); 
		        } 
		  
		        int nextInt() 
		        { 
		            return Integer.parseInt(next()); 
		        } 
		  
		        long nextLong() 
		        { 
		            return Long.parseLong(next()); 
		        } 
		  
		        double nextDouble() 
		        { 
		            return Double.parseDouble(next()); 
		        } 
		  
		        String nextLine() 
		        { 
		            String str = ""; 
		            try
		            { 
		                str = br.readLine(); 
		            } 
		            catch (IOException e) 
		            { 
		                e.printStackTrace(); 
		            } 
		            return str; 
		        } 
		    } 
		
		 //gcd of two number
		 
		    public static int gcd(int a, int b)
		    {
		      if (b == 0)
		        return a;
		      return gcd(b, a % b); 
		    }
		    
		 //--------------------------------------------------------------------------------------------
		 
		 //lcm of two number
		 static int x;
		    public static int lcm(int a, int b)
		    {
		        return (a / gcd(a, b)) * b;
		    }
		 
		  //-------------------------------------------------------------------------------------------
	
		   
		    
		    public static void main(String[] args) 
		    {
		    	
		    	FastReader s=new FastReader();
		       int t=s.nextInt();
		       while(t-->0)
		       {
		      int n=s.nextInt();
		      int[] a=new int[n];
		      
		      for(int i=0;i<n;i++)
		      {  a[i]=s.nextInt();}
		      
		      int[] b=new int[n];
		      Arrays.fill(b, -1);
		      
		      b[0]=a[0];
		      int count=a[0];
		      count--;
		      int flag=0;
		      for(int i=1;i<n;i++)
		      {
		    	  if(a[i]!=a[i-1])
		    		  count=a[i]-i;
		    	  
		    	  //System.out.println(count+" count");
		    	  count--;
		    	  if(count<0) {
		    		  flag=1;
		    		  break;}
		    		  
		      }
		      if(flag==1)
		    	  System.out.println("-1");
		      else 
		      {
		    	  int[] v=new int[n+1];
		    	 for(int i=1;i<n+1;i++)
		    		 v[i]=i;
		    	 int k=1;
		    	 System.out.print(a[0]+" ");
		    	 v[a[0]]=-1;
		    	 for(int i=1;i<n;i++)
		    	 {
		    		 if(a[i]==a[i-1])
		    		 {
		    			 while(v[k]==-1)
		    				 k++;
		    			 System.out.print(k+" ");
		    			 v[k]=-1;
		    		 }else 
		    		 {
		    			 System.out.print(a[i]+" ");
		    			 v[a[i]]=-1;
		    		 }
		    	 }
		    	  System.out.println();
		    	  
		    	  
		      }
		      
		      
		      
		      
		      
		      
		      
		       }
		    	
		    	
		    	
}}