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())

a = list(map(int,input().split()))
b = [-1 for i in range(n)]

for i in range(1,n):

	if a[i] != a[i-1]:
		b[i] = a[i-1]


MEX = 0
j = 0
# print(b)
l = []
while j <= n-1:

	if MEX < a[j]:
		l.append(MEX)
		MEX = MEX + 1
	elif MEX == a[j]:
		# print(j)
		MEX = MEX + 1
		j = j + 1
	elif MEX > a[j]:
		# print(MEX,"--")
		j = j + 1
	else:
		l.append(MEX)
		MEX = MEX + 1
while len(l) < n:
	l.append(MEX)
	MEX = MEX + 1

j = 0
for i in range(n):

	if b[i] == -1:
		b[i] = l[j]
		j = j + 1
print(*b)