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 atexit
import io
import sys
import math
from collections import defaultdict,Counter

# _INPUT_LINES = sys.stdin.read().splitlines()
# input = iter(_INPUT_LINES).__next__
# _OUTPUT_BUFFER = io.StringIO()
# sys.stdout = _OUTPUT_BUFFER

# @atexit.register
# def write():
#     sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())

# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')

# m=pow(10,9)+7
n=int(input())
a=list(map(int,input().split()))
b=[-1]*n
visit=[0]*1000001
visit[a[0]]=1
for j in range(1,n):
    if a[j]!=a[j-1]:
        b[j]=a[j-1]
    visit[a[j]]=1
start=0
for j in range(n):
    if b[j]==-1:
        while visit[start]==1:
            start+=1
        b[j]=start
        visit[start]=1
        start+=1
        if start>1000000:
            break
if b.count(-1)!=0:
    print(-1)
else:
    print(*b)