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.*;
import java.util.*;
public class EhabAndPrefixMEXs {
    public static void main(String agrs[])
    {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int[] arr = new int[n];
        int[] ans = new int[n];
        int[] counter = new int[100001];
        for(int i = 0; i < n; i++)
        {
            int temp = in.nextInt();
            counter[temp]++;
            arr[i] = temp;
        }
        Arrays.fill(ans, -1);
        for(int i = 0; i < n - 1; i++)
        {
            if(arr[i] < arr[i + 1])
            {
                ans[i + 1] = arr[i];
            }
        }
        int count = 0;
        for(int i = 0; i < n; i++)
        {
            if(ans[i] != -1)
            {
                continue;
            }
            while(counter[count] != 0)
            {
                count++;
            }
            counter[count]++;
            ans[i] = count;
        }
        for(int i = 0; i < n; i++)
        {
            if(i != n - 1)
            {
                out.print(ans[i] + " ");
            }
            else
            {
                out.println(ans[i]);
            }
        }
        out.close();
    }
}
