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())

arr = [int(i) for i in input().split()]

toAdd = set()
v = max(arr)
for i in range(v):
	toAdd.add(i)

# print(toAdd)

# print(*arr)

out = [-1]*n


if arr[0] == 1:
	out[0] = 0
	toAdd.remove(0)

for i in range(1,n):
	if arr[i-1] != arr[i]:
		out[i] = arr[i-1]
		toAdd.remove(arr[i-1])

# print(toAdd)

# print(out)

for i in range(n):
	if out[i]==-1:
		if len(toAdd) != 0:
			last = toAdd.pop()
		out[i] = last

print(*out)