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.InputStream;
import java.io.PrintStream;

public class ProblemC {

    private static int[] solve(int a[], int n) throws Exception {
        a[n + 1] = n + 1; // для удобства

        int[] b = new int[n + 2];
        // каждый раз решаем задачу на интервале от j..i, где А[i]==i, а j = i_предыдущее + 1
        int j = 1;
        for (int i = 1; i <= n + 1; i++) {
            if (a[i] == i) {
                b[j] = i - 1;
                for (int p = i; p > j; p--) {
                    b[p] = (a[p] > a[p - 1]) ? a[p - 1] : b[p + 1] + 1;
                }
                j = i + 1;
            }
        }
        return b;
    }

    public static void solve(InputStream in, PrintStream out) throws Exception {
        java.util.Scanner scanner = new java.util.Scanner(in);
        int n = scanner.nextInt();
        int[] a = new int[n + 2];
        for (int i = 1; i <= n; i++)
            a[i] = scanner.nextInt();
        int[] b = solve(a, n);
        for (int i = 1; i <= n; i++)
            out.print(b[i] + " ");
    }

    public static void main(String[] args) throws Exception {
        solve(System.in, System.out);
    }
}
