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.Arrays;
import java.util.Scanner;

public class C {

    public static void main(String[] args) {
        final Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();

        int[] b = new int[n];

        int neededUpTo = 0;
        boolean zeroUsed = true;

        for (int i = 0; i < n; i++) {
            int a = scanner.nextInt();

            if (i == 0 && a == 0) {
                neededUpTo = 1;
                zeroUsed = false;
            }

            if (!zeroUsed && a > 1) {
                b[i] = 0;
                zeroUsed = true;
                continue;
            }

            if (a - 1 > neededUpTo) {
                if (a - 1 > i) {
                    System.out.println(-1);
                    return;
                }

                for (int j = neededUpTo + 1, k = 0; j <= a - 1; j++, k++) {
                    b[i - k] = j;
                }

                neededUpTo = a - 1;
            } else {
                b[i] = neededUpTo;
            }
        }

        System.out.println(Arrays.toString(b));
    }
}
