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 EhabandPrefixMEXs {

	// https://codeforces.com/contest/1364/problem/C
	
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("EhabandPrefixMEXs"));

		int n = Integer.parseInt(in.readLine());  
		int[] a = new int[n];
		int[] b = new int[n];
		StringTokenizer st = new StringTokenizer(in.readLine());
		for (int i=0; i<n; i++) {
			a[i] = Integer.parseInt(st.nextToken()); 
			if (a[i] > i+1) {
				System.out.println(-1);
				return;
			}
		}
		PriorityQueue<Integer> need = new PriorityQueue<>();
		for (int i=n-1; i>=0; i--) {
			if (i == 0 && a[i] != 0) {
				b[i] = 0;
			}
			else if (i > 0 && a[i] > a[i-1]) {
				b[i] = a[i-1];
				for (int j=a[i-1]+1; j<a[i]; j++) {
					need.add(j);
				}
			}
			else {
				if (need.size() > 0) {
					b[i] = need.poll();
				}
				else {
					b[i] = (int)1e6;
				}
			}
		}
		for (int i=0; i<n; i++) {
			System.out.print(b[i] + " ");
		}
	}

}
