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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class C {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        StringBuilder sb = new StringBuilder("");
            String[] in = br.readLine().split(" ");
            int n = Integer.parseInt(in[0]);
            int arr[] = new int[n];
            in = br.readLine().split(" ");
            Set<Integer> set = new HashSet<>();
            for(int i=0;i<n;i++){
                arr[i]=Integer.parseInt(in[i]);
                set.add(arr[i]);
            }
            int b[] = new int[n];
            Arrays.fill(b, -1);
            for(int j=1;j<n;j++){
                if(arr[j]!=arr[j-1]){
                  b[j]  = arr[j-1];
                }
            }
            int curr = 0;
            for(int j=0;j<n;j++){
                if(b[j]==-1){
                    while(set.contains(curr)){
                        curr++;
                    }
                    b[j]=curr;
                    curr++;
                }
            }
            for(int j=0;j<n;j++){
                sb.append(b[j]);
                if(j==n-1){
                    sb.append("\n");
                }else{
                    sb.append(" ");
                }
            }

        out.print(sb);
        out.flush();
        br.close();
    }

public static long getAns(long n){
        if(n==0 || n==1){
            return n;
        }
        return n+getAns(n/2);
}
}

