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.ArrayDeque;
import java.util.Arrays;
import java.util.Scanner;

public class r649c {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		int[] a=new int[n];
		for(int i=0;i<n;i++) a[i]=scan.nextInt();
		
		ArrayDeque<Integer> q=new ArrayDeque<>();
		int[] res=new int[n];
		Arrays.fill(res,-1);
		int mex=0;
		
		for(int i=0;i<n;i++) {
			int dif=a[i]-mex;
			if(dif<0) {
				System.out.println(-1);
				return;
			}
			else if(dif==0) {
				q.offer(i);
			}
			else {
				if(q.size()<dif-1) {
					System.out.println(-1);
					return;
				}
				else {
					res[i]=mex++;
					int ptr=a[i]-1;
					for(int x=0;x<dif-1;x++) {
						int over=q.poll();
						res[over]=ptr--;
					}
				}
			}
		}
		while(!q.isEmpty()) {
			int x=q.poll();
			if(res[x]==-1) res[x]=1000000;
		}
		for(int i:res) System.out.print(i+" ");
	}
}