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.
# May the Speedforce be with us...
'''
problem statement achhe se padhna hai
age se bhi dekhna hai, piche se bhi dekhna hai
'''
#from math import gcd,ceil,sin,radians,cos
#from bisect import insort
#from collections import defaultdict as dd
def lcm(a,b):
	return (a*b)//gcd(a,b)
def inin():
	return int(input())
def inar():
	return list(map(int,input().split()))
def ar(element,size):
	return [element for i in range(size)]
def yes():
	print('Yes')
def no():
	print('No')
'''
def countDivisibles(A, B, M): #count numbers in the range[A,B] which are divisible by M
    if (A % M == 0): 
        return ((B // M) - (A // M)) + 1
    return ((B // M) - (A // M)) 
	
def  countSetBits(n): #count the number of set bits in the binary representation of an integer
    count = 0
    while (n): 
        count += n & 1
        n >>= 1
    return count
def digitsum(num): #return the sum of dogots of a number
	su=0
	while(num):
		su+=num%10
		num//=10
	return su
'''

Testcase=1
for _ in range(Testcase):
	n=inin()
	a=inar()
	i=1
	s=set(a[:])
	d=[-1 for i in range(n)]
	while(i<n):
		if a[i]!=a[i-1]:
			d[i]=a[i-1]
		i+=1
	a=set([i for i in range(0,100009)])
	a=a.difference(s)
	for i in range(n):
		if d[i]==-1:
			d[i]=a.pop()
	print(*d)
		