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 collections import Counter

input()
for sjdkla in range(1):
    arr = list(map(int, input().split()))
    count = Counter(arr)
    prev = arr[0]
    impossible = False
    for ind, i in enumerate(arr):
        if i > ind + 1:
            impossible = True
        break
    if impossible:
        print(-1)
        continue

    newarr = []
    newset = set()
    safe = 0
    newsafe = None
    while 1:
        if count[safe] == 0:
            break
        safe += 1
    for i in arr:
        if newsafe is not None:
            newarr.append(newsafe)
            newsafe = None
        else:
            newarr.append(safe)
            while 1:
                safe += 1
                if count[safe] == 0:
                    break
        count[i] -= 1
        if count[i] == 0:
            newsafe = i
    print(*newarr)