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.
# fmt: off
FAST_IO = 1
if FAST_IO:
    import io, sys, atexit
    read = iter(sys.stdin.read().splitlines()).__next__
    _OUTPUT_BUFFER = io.StringIO()
    sys.stdout = _OUTPUT_BUFFER
    @atexit.register
    def write():
        sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
else:
    read = input
readint = lambda: int(read())
readlist = lambda: list(map(int, read().split()))


def solve(N, A):
    free = []
    ans = [10**6] * N
    low = 0
    for i in range(N):
        free.append(i)
        while free and low < A[i]:
            j = free.pop()
            ans[j] = low
            low += 1
    return ans

N=readint()
A=readlist()
ans = solve(N,A)
if ans is None:
    print(-1)
else:
    print(' '.join(map(str, ans)))