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.PriorityQueue;
import java.util.Scanner;

public class ProblemC {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();	
		int[] a = new int[n];
		
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
		}
		
		PriorityQueue<Integer> free = new PriorityQueue<>();
		
		int[] b = new int[n];
		int curr = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] == curr) {
				free.add(i);
			} else {
				b[i] = curr;
				
				for (int j = curr + 1; j < a[i]; j++) {
					if (free.isEmpty()) {
						System.out.println(-1);
						return;
					}
					b[free.poll()] = j;
				}
				
				curr = a[i];
			}
		}
		
		while (!free.isEmpty()) {
			b[free.poll()] = n;
		}
		
		for (int i = 0; i < n; i++) {
			System.out.print(b[i] + " ");
		}
		System.out.println();
	}

}
