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.io.*;

public class Solution
{
	public static void main(String [] args) throws IOException
	{
		BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
			int n = Integer.parseInt(buf.readLine().trim());
			int a[] = new int[n];
			StringTokenizer in1 = new StringTokenizer(buf.readLine());
			int i = 0;
			while(in1.hasMoreTokens())
			{
				a[i++] = Integer.parseInt(in1.nextToken());
			}
			findSol(a);
	}
	public static void findSol(int a[])
	{
		int n = a.length;
		int b[] = new int[1000000];
		for(int i=n-1;i>0;i--)
		{
			b[a[i]] = -1;
		}
		int size = n;
		int i = 0;
		while(size > 0)
		{
			if(b[i] != -1)
			{
				System.out.print(i+" ");
				size--;
			}
			i++;
		}
			
		System.out.println();

	}
}
