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.*;
import java.lang.*;
import java.math.*;

public class C {
	public static void main(String[] args) throws Exception {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(bf.readLine());
        int[] arr = new int[n];
        StringTokenizer st = new StringTokenizer(bf.readLine());
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(st.nextToken());
        }
        int[] result = new int[n];
        LinkedList<Integer> required = new LinkedList<Integer>();
        int unused = arr[n-1] + 1;
        for (int i = n - 1; i > 0; i--) {
            if (arr[i] > arr[i-1]) {
                result[i] = arr[i-1];
                for (int j = arr[i-1] + 1; j < arr[i]; j++) {
                    required.add(j);
                }
            } else {
                if (required.size() == 0) {
                    result[i] = unused;
                } else {
                    result[i] = required.pop();
                }
            }
        }
        if (arr[0] == 0) {
            if (required.size() == 0) {
                result[0] = unused;
            } else {
                result[0] = required.pop();
            }
        } else {
            result[0] = 0;
        }
        StringBuilder sb = new StringBuilder();
        sb.append(result[0]);
        for (int i = 1; i < n; i++) {
            sb.append(' ');
            sb.append(result[i]);
        }
        out.println(sb.toString());
        out.close();
        System.exit(0);
	}
}