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.
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1364C extends PrintWriter {
	CF1364C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1364C o = new CF1364C(); o.main(); o.flush();
	}

	static final int A = 1000000;
	void main() {
		int n = sc.nextInt();
		int[] aa = new int[1 + n];
		for (int i = 1; i <= n; i++)
			aa[i] = sc.nextInt();
		int[] bb = new int[1 + n]; Arrays.fill(bb, -1);
		boolean[] used = new boolean[A + 1];
		for (int i = 1; i <= n; i++)
			if (aa[i - 1] < aa[i])
				used[bb[i] = aa[i - 1]] = true;
		used[aa[n]] = true;
		for (int a = 0, i = 1; i <= n; i++)
			if (bb[i] == -1) {
				while (used[a])
					a++;
				bb[i] = a++;
			}
		Arrays.fill(used, false);
		for (int a = 0, i = 1; i <= n; i++) {
			used[bb[i]] = true;
			while (used[a])
				a++;
			if (aa[i] != a) {
				println(-1);
				return;
			}
		}
		for (int i = 1; i <= n; i++)
			print(bb[i] + " ");
		println();
	}
}
