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.io.*;
import java.util.*;
public class EhabAndPrefixMEXs {
	private static StreamTokenizer st;
	private static int nextInt() throws IOException{
		st.nextToken();
		return (int)st.nval;
	}
	public static void main(String[] args) throws IOException{
		st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		int n = nextInt();
		int a[] = new int[n];
		HashSet<Integer>hs = new HashSet<>();
		for(int i = 0; i < n; ++i) {
			a[i] = nextInt();
			hs.add(a[i]);
		}
		int b[] = new int[n], pos = 0;
		boolean valid = true;
		for(int i = 0; i < n; ++i,++pos) {
			if(a[i]>i+1) {
				valid = false;
				break;
			}
			if(i!=0&&a[i-1]<a[i])b[i]=a[i-1];
			else {
				while(hs.contains(pos))++pos;
				b[i]=pos;
			}
		}
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		if(!valid)pw.println(-1);
		else {
			for(int i = 0; i < n-1; ++i)pw.print(b[i]+" ");
			pw.println(b[n-1]);
		}
		pw.close();
	}

}
