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
import math 
from decimal import *
def I(): return int(sys.stdin.readline())
def IL(): return list(map(int,sys.stdin.readline().strip().split()))
def ILS(): return list(map(str,sys.stdin.readline().strip().split()))

n = I()
a = IL()
b = [-1 for x in a]
flag = 1
m = {}
for i in range(n):
	m[a[i]] = i+1
for x in m :
	if m[x] < n :
		b[m[x]] = x

c =  []
for i in range(2*n+1):
	if i in m : continue
	c.append(i)

c.sort()
x = 0
i=0;j=0
while i<n and j<2*n :
	if b[i]!= -1 : 
		i+=1 
		continue
	b[i] = c[j]
	j+=1
	i+=1
check = [-1 for i in a]
temp =0 
mp = {}
# for i in range(2*n+1):
# 	mp[i] =
for i in range(n):
	mp[b[i]] = 1
	if b[i] == temp :
		while temp in mp and mp[temp] != 0:
			temp+=1
	check[i] = temp
# print(check)
# print(a)
# print(b)
for i in range(n):
	if check[i]!= a[i] :
		flag = 0 
		break
if flag == 1 :
	for bb in b : print(bb,end=" ")
	print()
else : print(-1)