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 heapq import heappush, heappop
n = int(input())
ar = list(map(int, input().split()))

valid = True
s = []
for i in range(n):
	if ar[i]>i+1:
		valid = False 
		break 
	else:
		if i>0:
			for j in range(ar[i-1]+1,ar[i]):
				heappush(s,j)
		elif ar[0]==1:
			heappush(s,0)
		# print(i,s)
if not valid:
	print(-1)
else:
	maxi = ar[-1]+1
	s_len = len(s)
	for i in range(s_len,n):
		heappush(s,maxi)
		maxi += 1
	# print(s)
	ans = []
	for i in range(n):
		k = heappop(s)
		ans.append(k)
		# s.remove(k)
		if i+1<n and ar[i]!=ar[i+1]:
			heappush(s,ar[i])
	print(*ans)