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.
ll = lambda: list(map(int, input().split()))
lls=lambda: list(map(str, input().split()))
st= lambda: input()
v = lambda: map(int, input().split())
ii = lambda: int(input())
#from math import *
from sys import *
from datetime import datetime
from sys import stdin, stdout
import sys
def lcm(a,b):
    return (a*b)//gcd(a,b)
from collections import *
def modFact(n):
    p=(10**9)+7
    if n >= p:
        return 0
    result = 1
    for i in range(1, n + 1):
        result = (result * i) % p
    return result
x=ii()
l=ll()
ans=[]
if(l[0]==0):
    ans.append(1)
    s=1
else:
    ans.append(0)
    s=0
for i in range(1,x):
    s+=1
    if(s not in l):
        ans.append(s)
    else:
        while(s in l):
            s+=1
        ans.append(s)
print(*ans)


