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.io.*;
public class Solution {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a[] = new int[n];
        for(int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        int impossible = 0;
        int flag = 0;
        for(int i = 0; i < n; i++) {
            if((a[i] - 1 > i)) {
                impossible = 1;
                break;
            }
            if(i > 0) {
                if(a[i] - 1 != a[i-1]) {
                    flag = 1;
                }
            }
        }
        if(impossible == 1) {
            System.out.println("-1");
        }
        else if(flag == 0) {
            for(int i= 0; i < n; i++) {
                if(i != n-1) System.out.print(i+" ");
                else System.out.println(i);
            }
        }
        else {
            int ans[] = new int[n];
            int temp[] = new int[n+1];
            int largest = n;
            temp[a[n-1]] = 1;
            for(int i = n-1; i >= 0; i--) {
                if(i > 0  && a[i] != a[i-1]) {
                    ans[i] = a[i-1];
                    temp[a[i-1]] = 1;
                }
                else {
                    while(temp[largest] == 1) {
                        largest -= 1;
                    }
                    ans[i] = largest;
                    temp[largest] = 1;
                    largest -= 1;
                }
            }
            for(int i = 0; i < n; i++) {
                if(i != n-1) System.out.print(ans[i]+" ");
                else System.out.println(ans[i]);
            }
        }
    }
}
