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 collections import defaultdict as dd
import math
import sys
input=sys.stdin.readline
def nn():
	return int(input())

def li():
	return list(input())

def mi():
	return map(int, input().split())

def lm():
	return list(map(int, input().split()))

n = nn()

nums = lm()
b = [10**6]*n
cur_min = 0
cur_place = 0
last = 0
bad= False
for i in range(len(nums)):
	#print(b)
	#print(cur_place)
	if nums[i]==last:
		continue
	else:
		b[i]=last
		cur_min+=1
	while cur_min<nums[i] and cur_place<i:
		if b[cur_place]==10**6:
			#print(cur_place)
			b[cur_place]=cur_min
			cur_min+=1
			cur_place+=1
		else:
			cur_place+=1
	#print(cur_min)
	if not cur_min==nums[i]:
		bad = True
		break
	else:
		last = nums[i]

if bad:
	print(-1)
else:
	print(*b)
