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 Main
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        ArrayList<Integer>a=new ArrayList<Integer>();
        ArrayList<Integer>b=new ArrayList<Integer>();
        for(int x=0;x<n;x++)
        {a.add(sc.nextInt());}
        int curr=-1;int i=0;
        for(int x=0;x<n;x++)
        {
            int temp=0;
            if(x!=0&&a.get(x-1)!=a.get(x))
            {temp=a.get(x-1);}
            else
            {
                while(i<n&&curr<a.get(n-1))
                {if(a.get(i)-curr<=0)
                {i++;}
                else
                {curr++;break;}}
                if(i==n)
                {curr=a.get(n-1)+2;}
                temp=curr;
            }
            b.add(temp);
        }
        for(int c:b)
        {System.out.print(c+" ");}
        System.out.println();
    }
}