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 sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
def li():
	return [int(x) for x in input().split()]
def fi():
	return int(input())
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())	
def gh():
	sys.stdout.flush()
def bo(i):
	return ord(i)-ord('a')
from copy import *
from bisect import *	

from itertools import permutations 



t=1

while t>0:
	t-=1
	n=fi()	
	a=li()
	d=[0 for i in range(n)]
	s=set([i for i in range(n+1)])
	p=s-set(a)
	s=list(s)
	s.sort()
	ff={}
	mex=0
	for i in range(n):
		if a[i]>i+1:
			print(-1)
			exit(0)
	j=0
	m=len(p)
	p=list(p)
	#print(p)		
	for i in range(n):
		if i>0 and a[i]!=a[i-1]:
			print(a[i-1],end=" ")
		elif j>=m :
			print(0,end=" ")
		else:
			print(p[j],end=" ")
			j+=1	
