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 qc{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int a[]=new int[n];
        int prev=-1;
        ArrayList<Integer> al=new ArrayList<Integer>();
        for(int i=0;i<n;i++){
            a[i]=sc.nextInt();
            if(a[i]!=prev){
                al.add(a[i]);
                prev=a[i];
            }
        }
        int curr=al.get(0);
        StringBuilder sb=new StringBuilder();
        TreeSet<Integer> ts=new TreeSet<>();
        int min=0;
        if(n==1){
            if(a[0]>0){
                System.out.println(0);
            }
            else{
                System.out.println(1);
            }
            System.exit(0);
        }
        int max=-1;
        int b[]=new int[n];
        Arrays.fill(b, -1);
        HashSet<Integer> hs=new HashSet<>();
        boolean x[]=new boolean[1000000];
        x[a[n-1]]=true;
        for(int i=n-1;i>=0;i--){
           // hs.add(a[i]);
            if(i>0 ){
                if(a[i]>a[i-1]){
                    b[i]=a[i-1];
                    x[a[i-1]]=true;
                    //max=Math.min(b[i],max);  
                }
                
            }
           

        }
        max=0;
        while(x[max])max++;
        for(int i=0;i<n;i++){
            if(b[i]==-1){
                b[i]=max;
                x[max]=true;
                while(x[max])max++;
            }
        }
        Arrays.fill(x,false);
        int mex=0;
        for(int i=0;i<n;i++){
            x[b[i]]=true;
            if(b[i]==mex){
                while(x[mex])mex++;
            }
            
            if(mex!=a[i]){
               // System.out.println(mex+" "+a[i]);
                System.out.println(-1);
                System.exit(0);
            }
           
            

        }
        
        for(int i=0;i<n;i++){
            sb.append(b[i]).append(" ");
        }
        System.out.println(sb);
            
        
    }
}