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.
def solve():
    n = int(input())
    arr = list(map(int, input().split(" ")))

    index = n - 1
    unique = set()

    while index >= 0:
        if arr[index] > index + 1:
            print(-1)
            return

        unique.add(arr[index])
        index -= 1

    ans = []
    if arr[0] == 1:
        ans.append(0)
        unique.add(0)
    else:
        current = 0
        while current in unique:
            current += 1
        ans.append(current)
        unique.add(current)

    queue = [i for i in range(n + 1) if i not in unique]
    prev = arr[0]
    l = 1

    while l < n:

        if arr[l] != prev and arr[l-1] in unique:
            unique.remove(arr[l-1])
            lowest = arr[l-1]
        else:
            lowest = queue.pop(0)

        ans.append(lowest)
        unique.add(lowest)

        prev = arr[l]
        l += 1

    print(*ans)
solve()