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 __future__ import print_function, division
from sys import stdin, stdout
from fractions import gcd
from math import *
from collections import *

rstr = lambda: stdin.readline().strip()
rstrs = lambda: [str(x) for x in stdin.readline().split()]
rint = lambda: int(stdin.readline())
rints = lambda: [int(x) for x in stdin.readline().split()]

n, a = int(input()), rints()
ans, a2, dis = [0] * n, deque(), set(a)
if a[0] == 0:
    ans = [1] * n

for i in range(2, a[-1] + 1):
    if i not in dis:
        a2.append(i)

for i in range(1, n):
    if a[i] != a[i - 1]:
        ans[i] = a[i - 1]
    else:
        if a2:
            ans[i] = a2.popleft()

print(' '.join(map(str, ans)))
