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 Main {

    static int mx = (int)1e7;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int n = Integer.parseInt(br.readLine());
        StringTokenizer st = new StringTokenizer(br.readLine());
        int[] arr = new int[n];
        HashSet<Integer> hs = new HashSet<>();
        for(int i = 0; i<n; i++){
            arr[i] = Integer.parseInt(st.nextToken());
            hs.add(arr[i]);
        }
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        for(int i = 1; i<n; i++){
            if(arr[i]!=arr[i-1]){
                ans[i] = arr[i-1];
            }
        }
        int num = 1;
        for(int i = 0; i<n; i++){
            if(ans[i]==-1){
                while(hs.contains(num) && ans[i]!=num) num++;
                ans[i] = num++;
            }
        }
        for(int z: ans) sb.append(z).append(" ");
        sb.append('\n');
        System.out.println(sb);
    }
}