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.
from sys import stdin


def inp():
    return stdin.buffer.readline().rstrip().decode("utf8")


def itg():
    return int(stdin.buffer.readline())


def mpint():
    return map(int, stdin.buffer.readline().split())


# ############################## import
# ############################## main
# for case in range(1, itg()+1):
def main():
    n = itg()
    arr = tuple(mpint())
    have_ans = all([(arr[i] <= arr[i+1] and arr[i] <= i+1) for i in range(n-1)])
    if not have_ans:
        return [-1]
    aa = set(arr)
    bb = sorted(set(range(n+7)) - aa, reverse=True)
    aa = sorted(aa, reverse=True)
    ans = [bb.pop()]
    for i in range(1, n):
        if arr[i-1] != arr[i]:
            ans.append(aa.pop())
        else:
            ans.append(bb.pop())
    return ans

print(*main())
# Please check!
