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.lang.*;
import java.io.*;
public class gfgcontest
 {
	public static void main (String[] args)
	 {
		Scanner input = new Scanner(System.in);
			int n = input.nextInt();
			HashSet<Integer> hs = new HashSet<>();
			int arr[] = new int[n];
			for(int i=0;i<n;i++) {
				arr[i] = input.nextInt();
				hs.add(arr[i]);
			}
			func(arr,n,hs);
		
	}
	static void func(int arr[],int n,HashSet<Integer> hs) {
		int ans[] = new int[n];
		Arrays.fill(ans,-1);
		if(arr[n-1]>n){
			System.out.println(-1);
			return;
		}
		for(int i=n-2;i>=0;i--) {
			if(arr[i]>i+1){
				System.out.println(-1);
				return;
			}
			if(arr[i+1]>arr[i]) {
				ans[i+1] = arr[i];
				hs.add(ans[i+1]);
			}
		}
		int val = 0;
		for(int i=0;i<n;i++) {
			if(ans[i]==-1) {
				while(hs.contains(val)) {
					val++;
				}
				ans[i] = val;
				hs.add(val);
			}
		}
		for(int i=0;i<n;i++) {
			System.out.print(ans[i]+" ");
		}
		System.out.println();
	}
 }