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.io.*;
import java.util.*;

public class Main {
    String fileName = "";

    public void solve() throws IOException {
        // code here
        boolean flag = true;
        int n = nextInt();
        int [] arr = new int[n];
        int count = 0;
        int max = 0;
        boolean [] used = new boolean[n];
        Arrays.fill(arr, -1);
        for (int i = 0; i < n; i++) {
            int a = nextInt();
            if (a > max) {
                count += (a - max - 1);
                max = a;
                arr[i] = a;
                used[a - 1] = true;
            } else {
                if (count > 0) {
                    count--;
                } else {
                    flag = false;
                }
            }
        }

        int x = 0;
        if (!flag) {
            out.println(-1);
        } else {
            for (int i = 0; i < n; i++) {
                if (arr[i] == -1) {
                    while (used[x]) {
                        x++;
                    }
                    arr[i] = x + 1;
                    x++;
                }
                out.print(arr[i] + " ");
            }
            out.println();
        }


    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new Main().run();
    }

    private void run() {
        try {
            if (fileName.equals("")) {
                br = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
            } else {
                br = new BufferedReader(new FileReader(fileName + ".in"));
                out = new PrintWriter(fileName + ".out");
            }

            int t = nextInt();
            for (int i = 0; i < t; i++) {
                solve();
            }

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

}