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.*;
import java.lang.*;

public class Solution {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer st = new StringTokenizer("");

    String next() {
        if (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {

            }
        }
        return st.nextToken();
    }

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

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

    public static void main(String[] args) {
        new Solution().solve();
    }

    void solve() {
        int t = 1;
        t = nextInt();
        for (int tt = 0; tt < t; tt++) {
            int n = nextInt();
            int flag = 0;
            int arr[] = new int[n];

            TreeSet<Integer> set = new TreeSet<>();
            for (int i = 0; i < n; i++) {
                arr[i] = nextInt();
                set.add(i + 1);
            }
            for (int i = 1; i < n; i++) {
                if (arr[i] < arr[i - 1]) {
                    flag = -1;
                    break;
                }
            }
            if (flag == -1)
                out.println(flag);
            else {
                int res[] = new int[n];
                res[0] = arr[0];
                set.remove(res[0]);
                for (int i = 1; i < n; i++) {
                    if (arr[i] != arr[i - 1]) {
                        res[i] = arr[i];
                        set.remove(res[i]);
                    }
                }
                for (int i = 0; i < n; i++) {
                    if (res[i] == 0) {
                        res[i] = set.first();
                        set.remove(res[i]);
                    }
                }
                for (int i = 1; i <= n; i++) {
                    set.add(i);
                }
                int max = 0;
                for (int i = 0; i < n; i++) {
                    if (set.contains(res[i]))
                        set.remove(res[i]);
                    max = Math.max(max, res[i]);
                    if (max != arr[i])
                        flag = -1;
                }
                if (set.size() == 0 && flag != -1) {
                    for (int i : res)
                        out.print(i + " ");
                    out.println();
                } else {
                    out.println(-1);
                }

            }
        }
        out.close();
    }
}