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 __future__ import division, print_function
import sys
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
#from functools import lru_cache   #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------

n = int(input())
a = list(mapi())

mp = defaultdict(int)
for i in a:
    mp[i]+=1
q = deque()
a = [0]+a
res = []
j =0
vis = defaultdict(int)
for i in range(n+1):
    if mp[i]:
        while mp[i]:
            if len(q)>0:
                mp[i]-=1
                tmp = q.popleft()
                res.append(tmp)
                vis[tmp]+=1
            else:
                while mp[j] or vis[j]:
                    j+=1
                vis[j]+=1
                res.append(j)
                j+=1
                mp[i]-=1
        q.append(i)
print(*res)