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 defaultdict
from collections import deque
import math
import heapq
import sys
import io, os
input = sys.stdin.readline
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from  bisect import *
rs = lambda: input().strip()
ri = lambda: int(input())
rl = lambda: list(map(int, input().strip().split()))
rls= lambda: list(map(str, input().split()))


n=int(input().strip())
a=rl()
s=list(set(a))
s.sort()
d=defaultdict(lambda:-1)
ans=[]
prev=0
for i in s:
    d[i]=1
if(a[0]==0):
    ans.append(1)
    prev=1
elif(a[0]==1):
    ans.append(0)
    prev=0
else:
    print(-1)
    exit()
if(d[0]==1 and d[1]==1):
    print(-1)
    exit()
f=1
d1=defaultdict(lambda :-1)
d2=defaultdict(lambda :-1)
for i in range(1,n):
    d1[a[i]]=1
    if(a[i]!=a[i-1]):
        ans.append(a[i-1])
        d2[ans[-1]]=1
        if(d1[a[i]]==1):
            print(-1)
            exit()
        continue
    while(1):
        prev=prev+1
        #print(i,prev)
        if(d[prev]==1):
            continue
        else:
            break
  #  print(i,prev)
    ans.append(prev)
    d1[prev]=1
    if (d1[a[i]] == 1):
        print(-1)
        exit()
    # if(a[i]-prev>1):
    #     f=0
    #     break
if(f):
    mx=0
    for i in range(0,n):
        mx=max(ans[i],mx)
        if(a[i]-mx>1):
            print(-1)
            exit()
    print(*ans)
else:
    print(-1)









