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.*;
public class C{
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        ArrayList<Integer> open = new ArrayList<>();
        int[] result = new int[n];
        int[] nums = new int[n];
        for(int i = 0; i < n; i++) {
            nums[i] = input.nextInt();
        }
        Arrays.fill(result, nums[n - 1] + 1);
        ArrayList<Integer> next = new ArrayList<>();
        for(int i = 0; i <= 1000000; i++) {
            next.add(i);
        }
        for(int i = 0; i < n; i++) {
            if(nums[i] != next.get(0)) {
                result[i] = next.remove(0);
            }
            int counter = 1;
            while(next.get(0) != nums[i]) {
                result[i - counter] = next.remove(0);
            }
        }
        for(int i = 0; i < n; i++) {
            if(result[i] == -1) {
                System.out.print((nums[n - 1] + 1) + " ");
            }
            else System.out.print(result[i] + " ");
        }
    }
}