Given an array a of length n, find another array, b, of length n such that:

  * for each i (1 ≤ i ≤ n) MEX(\\{b_1, b_2, …, b_i\})=a_i. 



The MEX of a set of integers is the smallest non-negative integer that doesn't belong to this set.

If such array doesn't exist, determine this.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i) — the elements of the array a. It's guaranteed that a_i ≤ a_{i+1} for 1≤ i < n.

Output

If there's no such array, print a single line containing -1.

Otherwise, print a single line containing n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ 10^6)

If there are multiple answers, print any.

Examples

Input


3
1 2 3


Output


0 1 2 

Input


4
0 0 0 2


Output


1 3 4 0 

Input


3
1 1 3


Output


0 2 1 

Note

In the second test case, other answers like [1,1,1,0], for example, are valid.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class C {
    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int n = in.nextInt();
        TreeSet<Integer> set = new TreeSet<>();
        for (int i = 0; i < n + n + n + n; i++) {
            set.add(i);
        }

        int arr[] = new int[n];
        int res[] = new int[n];
        int freq[] = new int[n + 1];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
            set.remove(arr[i]);
            freq[arr[i]]--;
        }

        for (int i = 0; i < n; i++) {
            res[i] = set.pollFirst();
            freq[arr[i]]++;
            if (freq[arr[i]] == 0) set.add(arr[i]);
        }

        for (int i : res) pw.print(i + " ");
        pw.println();

        pw.close();
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }
}