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.
from sys import stdin,stdout

# stdin = open("input.txt","r")
# stdout = open("output.txt","w")


n=int(stdin.readline().strip())
arr=list(map(int,stdin.readline().strip().split(' ')))
p=True
for i in range(len(arr)):
	if arr[i]>i+1:
		p = False
if not p:
	stdout.write("-1\n")
else:
	arr+=[arr[-1]+1]
	av=[1 for i in range(len(arr)+10)]
	tarr=[-1 for i in range(len(arr))]
	for i in range(1,len(arr)):
		if arr[i]!=arr[i-1]:
			tarr[i]=arr[i-1]
			av[arr[i-1]]=0

	ctr=0
	for i in range(len(tarr)):
		if tarr[i]==-1:
			while av[ctr]==0:
				ctr+=1
			tarr[i]=ctr;ctr+=1;

	stdout.write(' '.join(list(map(str,tarr[:-1])))+"\n")


