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.
# kartikay26

"""
https://codeforces.com/contest/1364/problem/C

Idea:
- whenever MEX increases it means that number was added
- when MEX increases by more than 1 it means the numbers
  in between were previously added
- create "slots" for numbers we do not know, then fill later
  with n+1 or the  numbers that we find later which should
  have come before
"""

def main():
    n = int(input())
    mex = [int(x) for x in input().split()]
    arr = [n+1] * n
    prev_mex = 0
    slots = []
    for i in range(n):
        if mex[i] > prev_mex:
            arr[i] = prev_mex
            nums_to_add = range(prev_mex+1,mex[i])
            for num in nums_to_add:
                if len(slots) == 0:
                    print(-1)
                    return
                slot = slots.pop()
                arr[slot] = num
        else:
            slots.append(i)
        prev_mex = mex[i]

    # print(arr)
    print(" ".join(str(x) for x in arr))

if __name__ == "__main__":
    main()

