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;
import java.util.Arrays;
import java.util.TreeSet;
import java.util.stream.IntStream;

public class ProblemC {

    private static int[] solve(int a[], int n) throws Exception {
        TreeSet<Integer> numbers = new TreeSet<>();
        IntStream.rangeClosed(0, n).forEach(numbers::add);
        numbers.remove(a[n]);
        // первый проход: встречаем разных соседей A[i-1]!=A[i] сразу B[i] = A[i-1]
        // внутри идём в обратку: B[j-1] = B[j]+1 пока B[j-1]<A[i]
        // последний проход - просто проставляем неиспользованные цифры, начиная с минимальной

        int[] b = new int[n + 1];
        Arrays.fill(b, -1);
        // каждый раз решаем задачу на интервале от j..i, где А[i]==i, а j = i_предыдущее + 1
        for (int i = 2; i <= n; i++) {
            if (a[i - 1] != a[i]) {
                b[i] = a[i - 1];
                numbers.remove(b[i]);
                for (int j = i, value = b[i] + 1; j > 1 && b[j] + 1 < a[i]; j--) {
                    if (b[j - 1] == -1) {
                        b[j - 1] = value;
                        numbers.remove(value);
                        value++;
                    }
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            if (b[i] == -1) {
                b[i] = numbers.first();
                numbers.remove(b[i]);
            }
        }
        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 + 1];
        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);
    }
}
