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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Set;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Kraken7
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    Scanner in = new Scanner(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    TaskC solver = new TaskC();
    solver.solve(1, in, out);
    out.close();
  }

  static class TaskC {

    public void solve(int testNumber, Scanner in, PrintWriter out) {
      int n = in.nextInt();
      int[] a = new int[n];
      boolean fine = true;
      Set<Integer> set = new HashSet<>();
      for (int i = 0; i < n; i++) {
        a[i] = in.nextInt();
        set.add(a[i]);
        if (a[i] > i + 1)
          fine = false;
      }
      if (!fine) {
        out.println(-1);
        return;
      }
      int[] b = new int[n];
      Arrays.fill(b, -1);
      for (int i = 1; i < n; i++) {
        if (a[i] != a[i - 1]) {
          b[i] = a[i - 1];
        }
      }
      int last = 0;
      for (int i = 0; i < n; i++) {
        if (b[i] != -1)
          continue;
        while (set.contains(last))
          last++;
        b[i] = last;
        last++;
      }
      for (int i : b)
        out.print(i + " ");
    }

  }
}

