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())
arr = list(map(int, input().split()))
max_el = arr[-1]
groups = [[arr[0], 1]]
has_ans = True
for i in range(1, n):
    if arr[i] > i + 1:
        has_ans = False
        break
    if arr[i] == groups[-1][0]:
        groups[-1][1] += 1
    else:
        groups.append([arr[i], 1])
if not has_ans:
    print(-1)
else:
    ans = [-1 for i in range(n)]
    last_group = groups.pop()
    i = 0
    for group in groups:
        i += group[1]
        ans[i] = group[0]
    groups.append(last_group)
    groups.append([2 * n, 0])
    val = 0 if arr[0] == 1 else 1
    group_i = 0 if arr[0] == 1 else 1

    for i in range(n):
        if ans[i] == -1:
            ans[i] = val
            val += 1
            while groups[group_i][0] == val:
                val += 1
                group_i += 1
    print(*ans)