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.
import sys
from array import array  # noqa: F401
import typing as Tp  # noqa: F401


def input():
    return sys.stdin.buffer.readline().decode('utf-8')


def output(*args):
    sys.stdout.buffer.write(
        ('\n'.join(map(str, args)) + '\n').encode('utf-8')
    )


def main():
    from heapq import heappop, heappush
    from collections import Counter
    n = int(input())
    a = list(map(int, input().split()))
    cnt = Counter(a)
    used = [0] * (2 * 10**5)
    hq = [i for i in range(2 * 10**5) if cnt[i] == 0]
    ans = []
    l = 0

    for i, v in enumerate(a):
        ans.append(heappop(hq))
        used[ans[-1]] = 1
        if not (all(used[j] for j in range(l, v)) and not used[v]):
            print(-1)
            exit()
        l = v
        cnt[v] -= 1
        if cnt[v] == 0:
            heappush(hq, v)

    output(' '.join(map(str, ans)))


if __name__ == '__main__':
    main()
