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.
//created by Whiplash99
import java.io.*;
import java.util.*;
public class C
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N,max=(int)(1e6);

        N=Integer.parseInt(br.readLine().trim());
        StringBuilder sb=new StringBuilder();

        int a[]=new int[N];
        int b[]=new int[N];

        String s[]=br.readLine().trim().split(" ");
        for(i=0;i<N;i++) a[i]=Integer.parseInt(s[i]);

        ArrayDeque<Integer> DQ=new ArrayDeque<>();
        int count=0;

        TreeSet<Integer> set=new TreeSet<>();
        for(i=0;i<=N+5;i++) set.add(i);

        if(a[0]==0)
        {
            b[0]=max;
            DQ.add(0);
        }
        else
        {
            b[0]=0; set.remove(0);
            count++;
        }

        for(i=1;i<N;i++)
        {
            if(a[i]!=a[i-1])
            {
                b[i] = a[i - 1]; set.remove(a[i-1]);
                count++;
            }
            else
            {
                b[i]=max;
                DQ.add(i);
            }

            if(a[i]>count)
            {
                int diff=a[i]-count;
                if(diff>DQ.size())
                {
                    System.out.println(-1);
                    System.exit(0);
                }
                while (!DQ.isEmpty())
                {
                    int pos=DQ.pollFirst();
                    b[pos]=set.pollFirst();
                    diff--;
                    if(diff==0) break;
                }
            }
        }
        for(i=0;i<N;i++) sb.append(b[i]).append(" ");
        System.out.println(sb);
    }
}