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.Scanner;

public class C1364 {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int length = scanner.nextInt();
		int[] numbers = new int[length];
		boolean[] used = new boolean[length + 1];
		int minimum = 0;
		for(int i = 0; i < length; i++) {
			int temp = scanner.nextInt();
			numbers[i] = temp;
			if(numbers[i] == minimum + 1)
				minimum++;
			used[temp] = true;
		}
		int[] answer = new int[length];
		int location = 0;
		while(used[location]) 
			location++;
		answer[0] = location;
		
		location++;
		
		for(int i = 1; i < length; i++) {
			if(numbers[i] == numbers[i - 1]) {
				
				while(used[location]) 
					location++;
				
				answer[i] = location;
				
				location++;
			}
			else {
				answer[i] = numbers[i - 1];
			}
		}
		for(int a: answer)
			System.out.print(a + " ");
	}
}