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.*;
import java.math.*;
public class Solution {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] ar = new int[n];
		int[] ans = new int[n];
		boolean flag = true;
		for(int i = 0; i < n; i++) ar[i] = in.nextInt();
		int cnt = 0;
		for(int i = 0; i < n; i++) {
			if(cnt == ar[n - 1]) cnt += 1;
			ans[i] = cnt;
			cnt += 1;
		}
		int last = n - 1;
		for(int i = n - 1; i >= 0; i--) {
			if(ar[i] - 1 > i) {
				flag = false;
				break;
			}else {
				if(ar[i] <= i) {
					if(ans[ar[i]] == ar[i]) {
						int temp = ans[ar[i]];
						ans[ar[i]] = ans[last];
						ans[last] = temp;
						last -= 1;
					}
				}
			}
		}
		if(flag == false) {
			System.out.println("-1");
		}else {
			for(int i = 0; i < n; i++) {
				if(i > 0) System.out.print(" ");
				System.out.print(ans[i]);
			}
			System.out.print("\n");
		}
	}
}