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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author anand.oza
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BBox solver = new BBox();
        solver.solve(1, in, out);
        out.close();
    }

    static class BBox {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int t = in.nextInt();
            for (int i = 0; i < t; i++) {
                solve(in, out);
            }
        }

        private void solve(InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int[] q = in.readIntArray(n);

            int[] p = new int[n];
            p[0] = q[0];
            TreeSet<Integer> left = new TreeSet<>();
            for (int i = 1; i <= n; i++)
                left.add(i);
            left.remove(q[0]);
            for (int i = 0; i + 1 < n; i++) {
                if (q[i + 1] > q[i]) {
                    p[i + 1] = q[i + 1];
                    left.remove(q[i + 1]);
                }
            }
            for (int i = 0; i < n; i++) {
                if (p[i] == 0) {
                    if (left.first() > q[i]) {
                        out.println(-1);
                        return;
                    }
                    p[i] = left.pollFirst();
                }
            }
            out.println(Util.join(p));
        }

    }

    static class Util {
        public static String join(int... i) {
            StringBuilder sb = new StringBuilder();
            for (int o : i) {
                sb.append(o);
                sb.append(" ");
            }
            if (sb.length() > 0) {
                sb.deleteCharAt(sb.length() - 1);
            }
            return sb.toString();
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

        public int[] readIntArray(int n) {
            int[] x = new int[n];
            for (int i = 0; i < n; i++) {
                x[i] = nextInt();
            }
            return x;
        }

    }
}