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.
# f = open('test.py')
# def input():
# 	return f.readline().replace('\n','')

# import heapq
# import bisect
# from collections import deque
from collections import defaultdict
def read_list():
	return list(map(int,input().split(' ')))
def print_list(l):
	print(' '.join(map(str,l)))

MAX = 9999999999
n = int(input())
a = read_list()
b = []
MEX = 0
dic = defaultdict(int)
for t in a:
	dic[t]+=1
m = 0
for i in range(n):
	if a[i]!=MEX:
		b.append(MEX)
		MEX = a[i]
		m = max(m,b[-1]+1)
	else:
		for tmp in range(m,MAX):
			if dic[tmp]==0:
				b.append(tmp)
				m = tmp+1
				break
		dic[a[i]]-=1
print_list(b)