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 Counter
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
import math

g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")

n, = gil()
a = gil()

flag = True
used = set([a[0]])
for i in range(1, n):
	used.add(a[i])
	if (i and a[i] < a[i-1]) or (a[i] > i+1): flag = False
 
if flag :
	h = []
	for i in range(int(1e6)+1):
		if i not in used : h.append(i)
	b = []
	heapify(h)
	a.append(1)
	for i in range(n):
		b.append(heappop(h) if h else b[-1])
		if a[i]!=a[i+1]:heappush(h, a[i])
	print(*b)
else:
	print(-1)
