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().strip().split()))
available_numbers = set(range(2*n))
removed = set()
for v in a:
    if v not in removed:
        available_numbers.remove(v)
        removed.add(v)
available_numbers = sorted(list(available_numbers))
mex = [-1] * n
for i, v in enumerate(a[1:]):
    if a[i] != v:
        mex[i+1] = a[i]


j = 0
for k, v in enumerate(mex):
    if v == -1:
        mex[k] = available_numbers[j]
        j+=1

print(" ".join(map(str, mex)))

"""
# Confirm mex
nums = list(range(2*n))[::-1]
removed = set()
listt = []
for v in mex:
    if v not in removed:
        nums.remove(v)
        removed.add(v)
    listt.append(nums[-1])

if listt == a:
    print(" ".join(map(str, mex)))
else:
    print("-1")
"""
