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.
n=int(input())
l=list(map(int,input().split()))
if max(l)>n:
	print(-1)
else:
	a=[]
	ans=[]
	for i in range(n+1):
		a.append(-1)
		ans.append(-1)
	for i in range(n):
		a[l[i]]=i+1
	for i in range(n+1):
		if a[i]!=-1:
			ans[a[i]]=i
	# for i in range(n+1):
	# 	print(a[i], ans[i])
	print()
	j=0
	for i in range(n+1):
		if ans[i]!=-1:
			continue
		while(a[j]!=-1):
			j+=1
		ans[i]=j
		# print(i,j,"t")
		j+=1
	for i in range(n):
		print(ans[i],end=" ")
