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 sys import stdin
input = stdin.buffer.readline

n = int(input())
*a, = map(int, input().split())
b = [-1] * n
vis = [0] * (n + 1)
x = y = 0
if a[0]:
	b[0] = 0
	vis[0] = 1
for i in range(1, n):
	if a[i] != a[i - 1]:
		b[i] = a[i - 1]
		vis[a[i - 1]] = 1
		while 1:
			while vis[y]:
				y += 1
			if y >= a[i]:
				break
			while b[x] != -1:
				x += 1
			b[x] = y
			vis[y] = 1
			x += 1

x = 100005
for i in range(n):
	if b[i] == -1:
		b[i] = x
		x += 1
print(*b)

