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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;

public class C {

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader file = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(file.readLine());
		boolean[] used = new boolean[1000001];
		int[] nums = new int[n];
		boolean[] last = new boolean[n];
		StringTokenizer st = new StringTokenizer(file.readLine());
		for(int i = 0; i < n; i++) {
			nums[i] = Integer.parseInt(st.nextToken());
		}
		boolean pos = true;
		for(int i = n-1; i >= 0; i--) {
			if(nums[i] > i+1)
				pos = false;
			if(!used[nums[i]])
				last[i] = true;
			used[nums[i]] = true;
		}
		if(!pos)
			System.out.println(-1);
		else {
			Queue<Integer> freed = new LinkedList<>();
			int index = 0;
			StringBuilder ans = new StringBuilder();
			for(int i = 0; i < n; i++) {
				while(used[index])
					index++;
				if(!freed.isEmpty()) {
					ans.append(freed.remove() + " ");
				}
				else {
					ans.append(index + " ");
					index++;
				}
			if(last[i])
				freed.add(nums[i]);
		}
		System.out.println(ans);
		}
		
	}

}
