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();
    int a[]=new int[n];
int b[]=new int[n];
int c[]=new int[1000001];
		for(int i=0;i<n;i++)
		{
			a[i]=sc.nextInt();
	
			if(i>0 && a[i]!=a[i-1])
			{
				b[i]=a[i-1];
				c[a[i-1]]=1;
			}
		}
		c[a[n-1]]=1;
		int ans=0;
		for(int i=0;i<n;++i)
		{
			if(b[i]==-1)
			{
				while(c[ans]!=0) 
				++ans;
				b[i]=ans;
				c[ans]=1;
				++ans;
			}
		}
		
		for(int i=0;i<n;++i)
System.out.print(b[i]+" ");
		System.out.println();
		
	
	
}
}
