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())
a = [int(x) for x in input().split(' ')]
a_dict = {}

for item in a:
    if item in a_dict.keys():
        a_dict[item] += 1
    else:
        a_dict[item] = 1

m = 0
ans = []
skip_print = False
for i in range(n):
    if a[i] > i + 1:
        print(- 1)
        skip_print = True
        break
    if i > 0 and a[i] != a[i - 1]:
        ans.append(a[i - 1])
    else:
        while m in a_dict.keys():
            m += 1
        ans.append(m)
        m += 1
if not skip_print:
    print(*ans)

