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:
// This template code suggested by KT BYTE Computer Science Academy
//   for use in reading and writing files for USACO problems.

// https://content.ktbyte.com/problem.java

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

public class Box {

    static StreamTokenizer in;

    static int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    public static void main(String[] args) throws Exception {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

        int t = nextInt();

        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

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

            int[] prefixSums = new int[n];

            prefixSums[0] = nextInt();

            ArrayList<Integer> changePoints = new ArrayList<>();

            for (int j = 1; j < n; j++) {
                prefixSums[j] = nextInt();

                if (prefixSums[j] != prefixSums[j-1]) changePoints.add(j);
            }

            int[] ans = new int[n];

            ans[0] = prefixSums[0];

            boolean[] used = new boolean[n+1];

            used[ans[0]] = true;

            int lastSure = 1;

            int last_change_point = 0;

            boolean fail = false;

            for (int j : changePoints) {
                ans[j] = prefixSums[j];
                used[ans[j]] = true;

                for (int k = last_change_point + 1; k < j; k++) {
                    while (used[lastSure]) {
                        lastSure++;
                    }

                    if (lastSure >= j) {
                        // FAIL - PRINT -1
                        fail = true;
                        break;
                    }

                    used[lastSure] = true;
                    ans[k] = lastSure;
                }

                last_change_point = j;

                if (fail) break;
            }

            // check from last change point to the end

            for (int k = last_change_point + 1; k < n; k++) {
                while (used[lastSure]) {
                    lastSure++;
                }

                if (lastSure >= ans[last_change_point]) {
                    // FAIL - PRINT -1
                    fail = true;
                    break;
                }

                used[lastSure] = true;
                ans[k] = lastSure;
            }

            if (fail) out.print(-1);

            else {
                for (int j = 0; j < n; j++) {
                    out.print(ans[j]);

                    if (j != n-1) out.print(" ");
                }
            }

            out.println();
        }

        out.close();
    }
}