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 C649_PC {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] a = new int[n];
		boolean[] taken = new boolean[1000005];
		int[] lastoccur = new int[1000005];
		for(int i =0;i<n;i++)
		{
			a[i]=in.nextInt();
			lastoccur[a[i]]=i;
			taken[a[i]]= true;
		}
		int[] ans = new int[n];
		Arrays.fill(ans, -1);
		for(int i =0;i<n;i++)
		{
			if(lastoccur[a[i]]!=n-1)
			{
				ans[lastoccur[a[i]]+1]=a[i];
			}
		}
		int index= 0;
		for(int i = 0;i<n;i++)
		{
			if(ans[i]==-1)
			{
				while(taken[index])
				{
					index++;
				}
				ans[i]= index;
				index++;
			}
		}
		taken = new boolean[1000005];
		index = 0;
		//System.out.println(Arrays.toString(ans));
		boolean works = true;
		for(int i =0;i<n;i++)
		{
			taken[ans[i]]=true;
			while(taken[index])
			{
				index++;
			}
			//System.out.println(a[i]+" "+index);
			if(a[i]!=index)
			{
				works = false;
				break;
			}
		}
		if(!works)
			System.out.println(-1);
		else
		{
			for(int i =0;i<n-1;i++)
				System.out.print(ans[i]+" ");
			System.out.println(ans[n-1]);
		}
		
	}

}
