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.*;
import java.io.*;

public class C {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        Hashtable<Integer, Integer> h = new Hashtable<Integer, Integer>(); 
  
        int n = Integer.parseInt(br.readLine());
        String[] line = br.readLine().split(" ");
        ArrayList<Integer> a = new ArrayList<Integer>();
        ArrayList<Integer> b = new ArrayList<Integer>();
        for (int i=0; i<n; i++) {
            a.add(Integer.parseInt(line[i]));
            h.put(a.get(i), i);
        }
 
        if (a.get(0) == 0)
            b.add(1);
        else 
            b.add(0);
        int c=1;

        for (int i=1;i<n;i++) {
            if (a.get(i) == a.get(i-1)) {
                c++;
                while (h.get(c) != null && h.get(c) >= i) 
                    c++;
                b.add(c);
            }
            else {
                c = a.get(i);
                b.add(a.get(i-1));
            }

        }
        for (int i=0; i<n; i++) 
            System.out.print(b.get(i) + " ");
    }
}