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.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class C {
	static ArrayList<Integer>[] grap;
	static String s2 = "Ashish", s1 = "Ayush";

	public static void main(String[] args) {

		Scanner nik = new Scanner(System.in);
		StringBuilder st = new StringBuilder();
		int n = nik.nextInt();
		int a[] = new int[n];

		int max = Integer.MIN_VALUE;
		for (int i = 0; i < n; i++) {
			a[i] = nik.nextInt();
			max = Math.max(max, a[i]);
		}

		int[] res = new int[n];

		if (max > n) {
			System.out.println(-1);
			return;
		}

		res[0] = a[0] == 0 ? 1 : 0;

		int idx = a[0] == 0 ? 0 : 1;
		HashSet<Integer> ht = new HashSet<>();
		ht.add(a[0]);
		for (int i = 1; i < n; i++) {

			if (a[i] != 0 && ht.contains(a[i - 1]) == false) {
				System.out.println(-1);
				return;
			}
			if (a[i] == idx) {
				res[i] = idx + 1;
				ht.add(idx + 1);
			} else {
				res[i] = idx;
				ht.add(idx);
				while (ht.contains(idx + 1)) {
					idx++;
				}

				idx++;
			}
		}

		for (int i = 0; i < n; i++) {
			st.append(res[i] + " ");
		}

		System.out.println(st);
	}

}