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.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        
        var sc = new Scanner(System.in);
    
        int n = Integer.parseInt(sc.next());
        var a = new int[n];
        var last = new int[n+1];
        Arrays.fill(last, -1);
        for(int i = 0; i < n; i++){
            a[i] = Integer.parseInt(sc.next());
            last[a[i]] = i;
        }
        var c = new int[n];
        Arrays.fill(c, -1);
        var pq = new PriorityQueue<Integer>();
        for(int i = 0; i <= n; i++){
            if(last[i] != -1){
                c[last[i]] = i;
            }else{
                pq.add(i);
            }
        }
        var sb = new StringBuilder();
        for(int i = 0; i < n; i++){
            if(!pq.isEmpty()){
                sb.append(pq.poll()).append(" ");
            }else{
                sb.append(100001).append(" ");
            }
            if(c[i] != -1){
                pq.add(c[i]);
            }
        }
        System.out.println(sb);
    }
}