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.BufferedReader;
import java.io.InputStreamReader;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class MEXFind {
    public static void main(String args[]) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(br.readLine());
        int[] a = new int[n];
        int[] dat = new int[2*n];

        StringTokenizer st = new StringTokenizer(br.readLine());
        for(int i = 0; i < n; i++){
            a[i] = Integer.parseInt(st.nextToken());
            dat[a[i]]++;
        }

        PriorityQueue<Integer> que = new PriorityQueue<>();
        for(int i = 0; i < 2*n; i++){
            if(dat[i] == 0) que.add(i);
        }

        int[] sol = new int[n];
        boolean valid = true;

        for(int i = 0; i < n; i++){
            sol[i] = que.poll();

            dat[a[i]]--;

            if(dat[a[i]] == 0){
                que.add(a[i]);
            }

            System.out.print(sol[i] + " ");
        }
    }
}
