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.awt.*;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;
import java.lang.String;

import static java.lang.Math.*;

public class Main {
    long mod = (long) 1e9 + 7;

    void run() throws IOException {
        int q = nextInt();
        while (q-- > 0) {
            int n = nextInt();
            int[] a = new int[n];
            for (int i = 0; i < a.length; i++) {
                a[i] = nextInt();
            }
            int[] ans = new int[n];
            ans[0] = a[0];
            TreeSet<Integer> ts = new TreeSet<>();
            for (int i = 0; i < n; i++) {
                ts.add(i + 1);
            }
            ts.remove(a[0]);
            boolean yes = false;
            for (int i = 1; i < a.length; i++) {
                if (a[i] == a[i - 1]) {
                    if (ts.lower(a[i]) == null) yes = true;
                    if (yes) {
                        pw.println(-1);
                        break;
                    }
                    ans[i] = ts.lower(a[i]);
                    ts.remove(ts.lower(a[i]));
                } else {
                    ans[i] = a[i];
                    ts.remove(a[i]);
                }
            }
            if (yes) continue;
            for (int i = 0; i < ans.length; i++) {
                pw.print(ans[i] + " ");
            }
            pw.println();
        }
        pw.close();
    }

    class Point {
        int x, y;

        public Point(int a, int b) {
            x = a;
            y = b;
        }
    }

    class PointComp implements Comparator<Point> {

        @Override
        public int compare(Point o1, Point o2) {
            if (o1.x == o2.x) return Integer.compare(o1.y, o2.y);
            return Long.compare(o1.x, o2.x);
        }
    }

    class Touple {
        int x;
        int from, to;

        public Touple(int a, int b, int c) {
            x = a;
            from = b;
            to = c;
        }
    }

    class ToupleComp implements Comparator<Touple> {


        @Override
        public int compare(Touple o1, Touple o2) {
            if (o1.x == o2.x && o1.from == o2.from) return Integer.compare(o1.to, o2.to);
            if (o1.x == o2.x) return Integer.compare(o1.from, o2.from);
            return Long.compare(o1.x, o2.x);
        }
    }

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");
    PrintWriter pw = new PrintWriter(System.out);

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

    String next() throws IOException {
        if (!st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

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

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

    public Main() throws FileNotFoundException {
    }

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


}