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.
def help():
	n = int(input())
	arr = list(map(int,input().split(" ")))

	ans_arr = [-1]*n
	dicti = {}

	if(arr[0]==1):
		ans_arr[0] = 0
		dicti[0]=1

	for i in range(1,n):
		if(arr[i] != arr[i-1]):
			ans_arr[i]=arr[i-1]
			dicti[arr[i-1]]=1
	dicti[arr[-1]]=1
	m = 0
	for i in range(n):
		while dicti.get(m,-1)==1:
			m+=1
		if(ans_arr[i]==-1):
			ans_arr[i]=m
			dicti[m]=1

	print(*ans_arr)

for _ in range(1):
	help()
