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 __future__ import print_function
import sys
if sys.subversion[0] == "PyPy":
    import io, atexit

    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))

    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()
RS = raw_input
RA = lambda x=int: map(x, RS().split())
RN = lambda x=int: x(RS())


def solve():
    if a != sorted(a):
        print(-1)
        return
    b = [-1]*n
    random_stack = [10**6]*(n+1)
    for i in reversed(range(1, n)):
        if a[i-1] == a[i]:
            b[i] = random_stack.pop()
        else:
            random_stack.extend(reversed(range(a[i-1], a[i])))
            b[i] = random_stack.pop()
    if a[0] == 0:
        b[0] = random_stack.pop()
    else:
        random_stack.append(0)
        b[0] = random_stack.pop()
    print(*b)



    return





n = RN()
a = RA()
solve()
