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.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        final Scanner in = new Scanner(System.in);

        final int n = in.nextInt();

        final int[] a = new int[n];
        boolean[] seen = new boolean[n + 1];

        for (int i = 0; i < n; i += 1) {
            a[i] = in.nextInt();
            seen[a[i]] = true;
        }

        int firstUnseen = n + 1;
        for (int i = 0; i <= n; i += 1) {
            if (!seen[i]) {
                firstUnseen = i;
                break;
            }
        }

        System.out.print(firstUnseen);
        for (int i = 1; i < n; i += 1) {
            if (a[i - 1] < a[i]) System.out.print(" " + a[i - 1]);
            else {
                for (int j = firstUnseen + 1; j <= n; j += 1) {
                    if (!seen[j]) {
                        firstUnseen = j;
                        break;
                    }
                }

                System.out.print(" " + firstUnseen);
            }
        }

        System.out.println();
    }
}
