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:
/*
ID: krishpa2
LANG: JAVA
TASK: b1227
*/ 

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

public class b1227 {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() throws IOException {
            while (st == null || !st.hasMoreElements()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }
        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        long nextLong() throws IOException {
            return Long.parseLong(next());
        }
        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
        String nextLine() throws IOException {
            return br.readLine();
        }
        void close() throws IOException {
            br.close();
        }
    }
    public static void main(String[] args) throws IOException{
        FastReader in = new FastReader(); 
        PrintWriter out = new PrintWriter(System.out); 

        int t = in.nextInt();
        while (t-- > 0) {
            int N = in.nextInt();
            int[] max = new int[N];
            int[] ans = new int[N];
            TreeSet<Integer> ts = new TreeSet<>();
            for (int i = 0; i<N; i++) {
                ts.add(i+1);
            }
            for (int i = 0; i<N; i++) {
                max[i] = in.nextInt();
            }
            ans[0] = max[0];
            ts.remove(ans[0]);
            boolean ok = true;
            for (int i = 1; i<N; i++) {
                if (max[i] != max[i-1]) {
                    ans[i] = max[i];
                    ts.remove(max[i]);
                }
                else if (ts.isEmpty() || ts.first() > max[i]) {
                    ok = false;
                }
                else {
                    ans[i] = ts.first();
                    ts.remove(ts.first());
                }
            }
            if (ok) {
                StringBuilder sb = new StringBuilder();
                for (int i: ans) {
                    sb.append(i).append(" ");
                }
                out.println(sb);
            }
            else {
                out.println(-1);
            }

        }

        in.close();
        out.close();
    }

}