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 = tuple(map(int, input().split()))
sa = set(a)
san = set()
min_not_in = 0
prev = 0
ans = [0] * n
for i in range(n):
    cur = a[i]
    if cur == prev:
        while min_not_in in sa:
            min_not_in += 1
        ans[i] = min_not_in
        sa.add(min_not_in)
        san.add(min_not_in)
    else:
        ans[i] = prev
        sa.add(prev)
        san.add(prev)
        while prev in san:
            prev += 1
        if cur != prev:
            print(-1)
            break
else:
    print(*ans)
