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 javax.swing.*;
import java.sql.ClientInfoStatus;
import java.util.*;

public class Example {
    static long[][] dp;
    //  static    int[]  ar;
    static int mod = 1000000000 + 7;
    static long[] ans = new long[100000 + 1];
    static List<List<Integer>> adj;
    static int[] vis;
    static int[] dis;
    static int row;
    static int col;
    static int[] arr;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t>0){
            t--;
            int n = sc.nextInt();
            int[] arr= new int[n];
            int[] frq= new int[n+1];
            for(int i=0;i<n;i++){
                arr[i]=sc.nextInt();
                frq[arr[i]]++;


            }
            Stack<Integer> list=new Stack<>();
            for(int i=1;i<=n;i++){
                if(frq[i]==0){
                    list.push(i);
                }
            }
            int k=0;

            boolean tt=true;
            for(int i=n-1;i>0;i--){
                if(arr[i-1]==arr[i]){
                    if(list.peek()>arr[i]){
                        System.out.println(-1);
                        tt=false;
                        break;
                    }
                    arr[i]=list.pop();

                }

            }
          if(tt){
              StringBuilder ans=new StringBuilder();
              for(int i=0;i<n;i++){
                  ans.append(arr[i]).append(" ");

              }
              System.out.println(ans);
          }
        }
    }




    public static int Check(int[] arr,int k){
        int sum=0;
        int j=1;
        int total=0;
        for(int i=1;i<arr.length;i++){
            sum+=arr[i];
            if(sum==k){
                total+=(i-j);
                j=i+1;
                sum=0;
            }if(sum>k){
                return arr.length;
            }
        }
        return total;


    }





    private static boolean answer(int i, int k) {
        if (k == 0)
            return true;
        if (k < 0)
            return false;
        if (i <= 0)
            return false;
        return answer(i, k - arr[i]) || answer(i - 1, k);
    }


    private static long findans(int[] arr, int i, int s) {
        if (i < 0 && s == 0)
            return 0;
        if (i < 0 && s == 1)
            return 0;
        if (dp[i][s] != -1) {
            return dp[i][s];
        } else {


            if (s == 0)
                return dp[i][s] = Math.max(findans(arr, i - 1, s), -1 * arr[i] + findans(arr, i - 1, 1 - s));
            else
                return dp[i][s] = Math.max(findans(arr, i - 1, s), arr[i] + findans(arr, i - 1, 1 - s));
        }
    }

    public static void dfs(int v, int di) {
        dis[v] = di;
        vis[v] = 1;
        for (int i = 0; i < adj.get(v).size(); i++) {
            int x = adj.get(v).get(i);
            if (vis[x] == 0) {
                dfs(x, di + 1);
            }
        }
    }

    private static boolean isPrime(long a) {
        if (a == 1)
            return true;

        for (int i = 2; i * i <= a; i++) {
            if (a % i == 0)
                return false;

        }
        return true;
    }

    private static int gcd(int aa, int bb) {
        if (bb == 0)
            return aa;
        return gcd(bb, aa % bb);

    }

    private static int lowerBound(int[] ar, int a) {
        int res = 0;
        int s = 1;
        int e = ar.length - 1;
        while (s <= e) {
            int mid = (s) + (e - s) / 2;
            if (ar[mid] <= a) {
                res = mid;
                s = mid + 1;

            } else {
                e = mid - 1;
            }
        }
        return res;
    }

    public static int getXORSum(int[] arr1, int[] arr2) {

        long c = 0;
        int n = arr1.length;
        int m = arr2.length;
        if (n == 1 && m == 1) {
            return arr1[0] & arr2[0];
        }
        int aa = n + m;
        for (int i = 0; i < 32; i++) {

            int z = 0;
            int o = 0;
            int cc = 0;
            for (int a : arr1) {


                int x = (a & (1 << i));
                if (x == 0)
                    z++;
                else
                    o++;


            }
            for (int a : arr2) {
                int x = (a & (1 << i));
                if (x == 0)
                    z++;
                else
                    o++;
            }

            System.out.println(z + " " + o);

            if (z != aa && o != aa) {
                c = c + (1 << i);

            }


        }
        return (int) c;


    }


}