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 = list(map(int, input().split()))


def ehab(n, a):
    set_a = set(a)
    b = []
    ans = []
    for i in range(n+1):
        if i not in set_a:
            b.append(i)
    index_b = 0
    ans.append(b[0])
    index_b += 1

    for i in range(1, n):
        if a[i] != a[i-1]: #kalo berubah
            ans.append(a[i-1])
        else:
            ans.append(b[index_b])
            index_b += 1
    return ans

print(ehab(n, a))



# def ehab(n, a):
#     ds = BukaTutup()

#     for a_num in a:
#         ds.tutup(a_num)

#     for a_num in a:
#         b.append(ds.mex())
#         ds.tutup(ds.mex())
#         ds.buka(a_num)

#     print(" ".join(b))


# class BukaTutup(object):
#     def __init__(self):
#         pass


#     def tutup(self, num):  # bisa beberapa lapis
#         pass


#     def buka(self, num):  # bisa beberapa lapis
#         pass


#     def mex(self, num):
#         pass
