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 sys import stdin, setrecursionlimit, stdout
#setrecursionlimit(1000000)
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline().rstrip()
def lsi(): return list(si())
#mod=1000000007
res=['Yes', 'No']
############# CODE STARTS HERE #############

test_case=1
while test_case:
    test_case-=1

    n=ii()
    a=li()
    b=[-1]*n
    v=[0]*(2*n+3)
    for i in range(1, n):
        if a[i]!=a[i-1]:
            b[i]=a[i-1]
        v[a[i]]=1
    v[a[0]]=1
    j=0
    for i in range(n):
        if b[i]==-1:
            while v[j]:
                j+=1
            b[i]=j
            j+=1
    print(*b)