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:
/*input
4
5
1 3 4 5 5
4
1 1 3 4
2
2 2
1
1


*/



import java.math.*;
import java.io.*;
import java.util.*;
public class Main
{
    static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead;
public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        }
public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        }
public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        }
public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9');
if (neg) 
                return -ret; 
            return ret; 
        }
public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        }
public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read();
do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9');
if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            }
if (neg) 
                return -ret; 
            return ret; 
        }
private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        }
private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        }
public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 




        
    static Reader sc=new Reader();
    static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));


static boolean flag=false;
    
    public static void main(String args[])throws IOException
    {
        /*
         * For integer input: int n=inputInt();
         * For long input: long n=inputLong();
         * For double input: double n=inputDouble();
         * For String input: String s=inputString();
         * Logic goes here
         * For printing without space: print(a+""); where a is a variable of any datatype
         * For printing with space: printSp(a+""); where a is a variable of any datatype
         * For printing with new line: println(a+""); where a is a variable of any datatype
            Scanner in = new Scanner(System.in);
        //all four
         int[] dr = { 1, 0, -1, 0 };
         int[] dc = { 0, 1, 0, -1 };
         println("Case #"+k+": "+1+"");//Google

         */
          Scanner in = new Scanner(System.in);
        int t =inputInt();
      
            while(t-->0)
            {
                StringBuilder sb = new StringBuilder();
                StringBuilder sb1 = new StringBuilder();
                
                  int n=inputInt();
                  int flag=0;
                  HashSet<Integer> hs  = new HashSet<>();
                  int small=1;

                 
                 for(int i=0;i<n;i++)
                 {
                    int x=inputInt();
                    if(!hs.contains(x))
                    {
                        if(small==x)
                            small++;

                        sb.append(x).append(" ");
                        hs.add(x);
                    }
                    else
                    {
                        while(hs.contains(small) && small<=n)
                            small++;

                        if(small>n || small>=x)
                        {
                            flag=1;
                           
                        }
                        sb.append(small).append(" ");
                        hs.add(small);

                    }
                 }

                 if(flag==1)
                    println(-1+"");
                else
                    println(sb+"");

            }

                      
            
                    

                    
            

            



    
bw.flush();
  bw.close();

}


           


static void dfs(int src,int dst,List<List<Integer>> ls,boolean visit[])
    {
        
        if(visit[src])
            return;
        visit[src]=true;
        
        if(src==dst)
            {flag=true;return;}

        Iterator<Integer> ite = ls.get(src).listIterator();

        while(ite.hasNext())
        {
            int x=ite.next();

            if(!visit[x])
            {
                
                dfs(x,dst,ls,visit);
            }
        }
        
    }

        public static long pow(long a,long b)
        {
            long result=1;

            while(b>0)
            {
                if(b%2==1) result*=a;
                a*=a;
                b/=2;
            }
            return result;
        }

        public static long nCr(long n, long r) 
        { 
            long ans=1;
            if(r>n-r)
            {
                r=n-r;
            }
            for(long i=1;i<=r;i++)
            {
                ans*=(n-i+1);
                ans/=i;
            }
            return ans;
        } 
        public static long modulo(long x,long N)
        {
            return (x % N + N) %N;
        }
        public static long lcm(long a,long b)
        {    
            return a / gcd(a, b) * b;
        }
        public static long gcd(long a,long b)
        {
            if(b==0)
                return a;
            return gcd(b,a%b);
        }

     public static int inputInt()throws IOException
    {
        return sc.nextInt();
    }
    public static long inputLong()throws IOException
    {
        return sc.nextLong();
    }
    public static double inputDouble()throws IOException
    {
        return sc.nextDouble();
    }
   
 
    public static String inputString()throws IOException
    {
        return sc.readLine();
    }
    public static void print(String a)throws IOException
    {
        bw.write(a);
    }
    public static void printSp(String a)throws IOException
    {
        bw.write(a+" ");
    }
    public static void println(String a)throws IOException
    {
        bw.write(a+"\n");
    }
}