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.lang.*;
import java.io.*;
 
public class C
{
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        String[] line = br.readLine().split(" ");
        int[]a=new int[n+1];
        int[]b=new int[n+1];
        int[]h=new int[n+1];
        
        for(int i=1;i<=n;i++) {
            a[i]=Integer.parseInt(line[i-1]);
            h[a[i]]=i;
        }
        
        int c=0;
        while (h[c]>0) c++;
        
        int next=-1;
        
        for (int i=1;i<=n;i++) {
            if (next==-1) { b[i]=c; c++; }
            else b[i]=next;
            
            if (h[a[i]]==i) next=a[i];
            else {
                while (c<=n && h[c]>0) c++;
                next=-1;
            }
        }
        
        for (int x=1;x<=n;x++) System.out.print(b[x]+" ");
    }

        
    }
    
