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
### START FAST IO ###
os_input = os.read(0, int(1e7)).split()
os_input_pos = -1
answer_list = []
def read_s():
    global os_input_pos
    os_input_pos += 1
    return os_input[os_input_pos].decode()
def read_i():
    return int(read_s())
def write_s(v):
    answer_list.append(v)
def write_i(v):
    write_s(str(v))
def print_ans():
    os.write(1, "\n".join(answer_list).encode())
#### END FAST IO ####

n = read_i()
a = [read_i() for i in range(n)]
need = [i != a[-1] for i in range(n+1)]
ans = [-1 for i in range(n)]
for i in range(n):
    if i > 0 and a[i] != a[i-1]:
        ans[i] = a[i-1]
        need[ans[i]] = False

j = 0
for i in range(n):
    if ans[i] == -1:
        while not need[j]:
            j += 1
        ans[i] = j
        need[j] = False

write_s(' '.join([str(x) for x in ans]))
print_ans()

