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.Reader;
import java.util.StringTokenizer;

public class TaskB {
    public static String doMain(Reader reader) throws IOException {
        MyReader in = new MyReader(reader);
        int i = in.nextInt();
        StringBuilder sb = new StringBuilder();
        for (int j = 0; j < i; j++) {
            sb.append(task(in));
            sb.append("\n");
        }
        return sb.toString();
    }

    private static String task(MyReader in) throws IOException {
        int n = in.nextInt();
        int[] q = new int[n];
        for (int i = 0; i < n; i++) {
            q[i] = in.nextInt();
        }
        int[] res = new int[n];
        boolean[] got = new boolean[n + 1];

        res[0] = q[0];
        got[res[0]] = true;
        int last = 1;
        for (int i = 1; i < n; i++) {
            if(q[i] < q[i-1] || q[i] > n) return "-1";
            if(q[i] > q[i-1]) {
                res[i] = q[i];
                got[res[i]] = true;
            } else {
                while(last < i + 1 && got[last]) ++last;
                if(last == i + 1) return "-1";
                res[i] = last++;
            }
        }
        String prefix = "";
        StringBuilder sb = new StringBuilder();
        for (int i : res) {
            sb.append(prefix);
            prefix = " ";
            sb.append(i);
        }
        return sb.toString();
    }


    public static void main(String[] args) throws IOException {
        String result = doMain(new InputStreamReader(System.in));
        System.out.println(result);
    }

    static class MyReader {
        BufferedReader bf;

        StringTokenizer st;

        String last;

        MyReader(Reader reader) throws IOException {
            bf = new BufferedReader(reader);
            readNextLine();
        }

        String nextToken() throws IOException {
            while (!st.hasMoreTokens()) {
                readNextLine();
            }
            return st.nextToken();
        }

        void readNextLine() throws IOException {
            last = bf.readLine();
            if (last == null) last = "";
            st = new StringTokenizer(last);
        }

        String nextLine() throws IOException {
            String s = last;
            readNextLine();
            return s;
        }

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

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

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

        int[] readIntArray(int n) throws IOException {
            int[] answer = new int[n];
            for (int i = 0; i < n; ++i) {
                answer[i] = nextInt();
            }
            return answer;
        }

        long[] readLongArray(int n) throws IOException {
            long[] answer = new long[n];
            for (int i = 0; i < n; ++i) {
                answer[i] = nextLong();
            }
            return answer;
        }

        double[] readDoubleArray(int n) throws IOException {
            double[] answer = new double[n];
            for (int i = 0; i < n; ++i) {
                answer[i] = nextDouble();
            }
            return answer;
        }
    }
}