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.Arrays;
import java.util.Scanner;

public class C_Ehab_and_Prefix_MEXs {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt(), c = 0;
		int[] n = new int[t + 1];
		int[] m = new int[100005];
		int[] o = new int[100005];
		Arrays.fill(m, -1);
		for(int i = 1; i <= t; i++) {
			n[i] = sc.nextInt();
			if(i > 1 && n[i] > n[i - 1]) {
				m[i] = n[i - 1];
				o[m[i]] = 1;
			}
		}
		o[n[t]] = 1;
		for(int i = 1; i <= t; i++) {
			while(o[c] == 1) c++;
			if(m[i] == -1) {
				m[i] = c;
				o[c] = 1;
			}
			System.out.print(m[i] + " ");
		}
		sc.close();
	}

}
