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.HashSet;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;

public class mex {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        StringBuffer sb = new StringBuffer();
        int n = sc.nextInt();
        int ar[] = new int[n];
        Set<Integer> s1=new HashSet<>();

        for (int i = 0; i < n; i++) {
            ar[i] = sc.nextInt();
            s1.add(ar[i]);
        }
        int ls = ar[n - 1];
        Stack<Integer> s = new Stack<>();
        if (ls > n) {
            sb.append("-1\n");
        }
        else
        {
            int c=ar[0]-1<0?1:0;
            int f=ar[0];
            int ls2=c-1;
            for (int i = 0; i < n; i++) {


                if(ar[i]==f)
                {
                    c=ls2+1;
                    while(s1.contains(c))
                    {
                        c++;
                    }

                    sb.append(c+" ");
                    s1.add(c);
                    ls2=c;
                }
                else
                {
                    c=ar[i-1];
                    sb.append(c+" ");
                    f=ar[i];
                }
            }
            // sb.append(s.size() + "\n");
            for (int i : s) {
                sb.append(i + " ");
            }
        }
        sb.append("\n");
        System.out.println(sb.toString());
    }
}

