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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class nB {

    public void run() throws NumberFormatException, IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        final int t = nextInt();
        for (int i = 0; i < t; i++) {
            boolean flag = false;
            int[] res = new int[nextInt()];
            int[] q = new int[res.length];
            boolean[] p = new boolean[res.length];
            for (int j = 0; j < q.length; j++) {
                q[j] = nextInt() - 1;
                if ((j > 0 && q[j] < q[j - 1]) || q[j] < j || q[j] >= q.length) {
                    flag = true;
                }
                if (j == 0 || (q[j] > q[j-1])) {
                    res[j] = q[j];
                    p[res[j]] = true;
                } else {
                    res[j] = -1;
                }
            }
            int ind = 0;
            for (int j = 0; j < res.length; j++) {
                if (res[j] == -1) {
                    while (p[ind]) {
                        ind++;
                    }
                    res[j] = ind;
                    ind++;
                }
            }
            if (flag) {
                pw.print(-1);
            }else {
                for (int x : res) {
                    pw.print(x + 1 + " ");
                }
            }
            pw.println();
        }
        pw.close();
    }

    StringTokenizer st;
    BufferedReader br;
    PrintWriter pw;

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

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

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


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

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