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 os,io
from sys import stdout
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n=int(input())
a=list(map(int,input().split()))
v1=[False]*(n+1)
s1=sorted(list(set(a)),reverse=True)
for i in s1:
    v1[i]=True
tmp=0
b=[0]*n
flag=True
for i in range(n):
    if len(s1)>0 and a[i]>s1[-1]:
        b[i]=s1[-1]
        s1.pop()
    elif flag==True:
        if tmp==a[i] or v1[tmp]==True:
            while True:
                tmp+=1
                if tmp>n:
                    flag=False
                    b[i]=n+5
                    break
                if v1[tmp]==False:
                    b[i]=tmp
                    tmp+=1
                    break
        else:
            b[i]=tmp
            tmp+=1
    else:
        b[i]=n+5
print(*b)