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.
n = int(input())
data = list(map(int, input().split()))

no = False

for i in range(n):
    if i + 1 < data[i]:
        no = True

my_set = set()

if no:
    print(-1)
else:
    for x in data:
        my_set.add(x)
    count = 0
    # Start
    while count in my_set:
        count += 1
    print(count, end=' ')
    my_set.add(count)
    # Then
    for i in range(1, n):
        # deterministic
        if data[i] != data[i - 1]:
            print(data[i - 1], end=' ')
            continue
        while count in my_set:
            count += 1
        print(count, end=' ')
        my_set.add(count)
    print()