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.
import math
from collections import defaultdict, Counter, deque

INF = float('inf')

def gcd(a, b):
	while b:
		a, b = b, a%b
	return a


def primeFactor(n):
	if n % 2 == 0:
		return 2
	i = 3
	while (i ** 2) <= n:
		if n % i == 0:
			return i 
		i += 1
	return n

def main():
	n = int(input())
	arr = list(map(int, input().split()))

	ans = [0 for i in range(n)]
	smallest = []
	s = 0
	while s < arr[0]:
		smallest.append(s)
		s += 1
	s = arr[0] + 1
	if n == 1:
		smallest.append(s)
	for i in range(1, n):
		while s < arr[i]:
			smallest.append(s)
			s += 1
		s = arr[i] + 1
		if i == n - 1:
			smallest.append(s)
	j = 0
	
	for i in range(n):
		if arr[i] > arr[i-1]:
			ans[i] = arr[i - 1]
		else:
			if j < len(smallest):
				ans[i] = smallest[j]
				last = smallest[j]
				# last += 1
			else:
				ans[i] = last
				last += 1
			j += 1

	print(*ans)
	# print(smallest)
if __name__ == "__main__":
	# t = int(input())
	t = 1
	for _ in range(t):
		main()







