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.
input()
maybe_used = []

data = list(map(int, input().split()))
result = [None] * len(data)

if data[0] == 0:
	maybe_used.append(0)
elif data[0] == 1:
	result[0] = 0
else:
	print(-1)

break_flag = False

for i in range(1, len(data)):
		if data[i] == data[i-1]:
			maybe_used.append(i)
		elif data[i] == data[i-1] + 1:
			result[i] = data[i-1]
		else:
			result[i] = data[i-1]
			for j in range(data[i-1]+1, data[i]):
				if maybe_used:
					k = maybe_used.pop(0)
					result[k] = j
					pass
				else:
					break_flag = True
					break
			if break_flag:
				print(-1)
				break

if not break_flag:
	x = 100500
	while maybe_used:
		result[maybe_used.pop(0)] = x
		x += 1
	print(*result)
	pass