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 heapq

n = int(input())
*arr, = map(int, input().split())

first = {}
last = {}

for i in range(n):
    last[arr[i]] = i

big_number = max(arr) + 1

b = [-1] * n

used = set()
for i in range(n):
    if i == 0 or arr[i] != arr[i - 1]:
        last_idx = last[arr[i]]
        if last_idx != n - 1:
            b[last_idx + 1] = arr[i]
            used.add(arr[i])
used.add(arr[-1])

max_val = max(arr) + 1
l = arr[-1] - 1

l = 0

for i in range(n):
    if b[i] == -1:
        while l in used:
            l += 1
        b[i] = l
        l += 1


print(*b)
