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.util.*;
import java.io.*;
//Captain on duty!



public class Main {

    static void compare(Main.pair a[] , int n) {

        Arrays.sort(a, new Comparator<Main.pair>() {
            @Override
            public int compare(Main.pair p1, Main.pair p2) {
                return p1.f - p2.f;
            }
        });
    }
    public static boolean checkPalindrome(String s)
    {
        // reverse the given String
        String reverse = new StringBuffer(s).reverse().toString();

        // check whether the string is palindrome or not
        if (s.equals(reverse))
            return true;

        else
            return false;
    }

    static class pair implements Comparable
    {
        int f;
        int s;
        pair(int fi, int se)
        {
            f=fi;
            s=se;
        }
        public int compareTo(Object o)//desc order
        {
            pair pr=(pair)o;
            if(s>pr.s)
                return -1;
            if(s==pr.s)
            {
                if(f>pr.f)
                    return 1;
                else
                    return -1;
            }
            else
                return 1;
        }
        public boolean equals(Object o)
        {
            pair ob=(pair)o;
            if(o!=null)
            {
                if((ob.f==this.f)&&(ob.s==this.s))
                    return true;
            }
            return false;
        }
        public int hashCode()
        {
            return (this.f+" "+this.s).hashCode();
        }


    }

    public static boolean palin(int l, int r, char[] c) {
        while (l <= r) {
            if (c[l] != c[r]) return false;
            l++;
            r--;
        }
        return true;
    }


    public static long gcd(long a, long b)
    {
        if(b==0)
            return a;

        return gcd(b, a%b);
    }

    public static long hcf(long a, long b)
    {
        long t;
        while(b != 0)
        {
            t = b;
            b = a%b;
            a = t;
        }
        return a;
    }

    public static boolean isPrime(int n)
    {
        if (n <= 1)
            return false;

        // Check from 2 to n-1
        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    public static String reverse(String str)
    {
        String str1="";
        for(int i=0;i<str.length();i++)
        {
            str1 = str1 + str.charAt(str.length()-i-1);
        }
        return str1;
    }



    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }


    }


    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];
            int[] vis=new int[100001];
            HashSet<Integer> h=new HashSet<>();
            for(int i=0;i<n;i++)
            {
                a[i]=s.nextInt();
                h.add(a[i]);
            }
            int[] l=new int[n];
            int c=0;
            for(int i=0;i<n;i++)
            {
                if(!h.contains(i+1))
                {
                    l[c]=i+1;
                    c++;
                }

            }
            int[] ans=new int[n];
             c=0;
            ans[0]=a[0];
            int flag=0;
            for(int i=1;i<n;i++)
            {
                if(a[i]>a[i-1])
                {
                    ans[i]=a[i];
                }
                else
                {
                    int temp = l[c];
                    c++;
                    if(temp<a[i])
                        ans[i]=temp;
                    else
                    {
                        flag=1;
                        break;
                    }
                }
            }
            if(flag==1)
                System.out.println(-1);
            else
            {
                for(int i=0;i<n;i++)
                    System.out.print(ans[i]+" ");
                System.out.println();
            }




        }
    }
}