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 functools import lru_cache
from sys import stdin, stdout
import sys
from math import *
# sys.setrecursionlimit(3*int(1e5))
# input = stdin.buffer.readline
# print = stdout.write
 
# @lru_cache()

n=int(input())
ar=list(map(int,input().split()))
ans=[]
for i in range(n):
	if(ar[i]>i+1):
		print(-1)
		exit()

mx=[0]*(n)
mx[-1]=inf
for i in range(n-2,-1,-1):
	if(ar[i+1]>ar[i]):
		mx[i]=ar[i+1]
	else:
		mx[i]=mx[i+1]

		
if(ar[0]==0):
	if(mx[0]==1):
		ans.append(max(ar)+1)
	else:
		ans.append(1)
	l=0
elif(ar[0]==1):
	ans.append(0)
	l=1

# for i in range(1,n):
# print(ans)
# print(ar)
for i in range(1,n):
	if(ar[i]==ar[i-1]):
		if(ans[i-1]+1<mx[i]):
			if(ar[i]==ans[i-1]+1):
				ans.append(ar[i]+1)
			else:
				ans.append(ans[i-1]+1)
		else:
			ans.append(ans[i-1])
	else:
		ans.append(l)
		l=ar[i]
print(*ans)





