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()))

alphabet = list()
alphabet.append(a[0])

for i in range(1, n):
    if a[i] > i + 1:
        print(-1)
        exit()

    if a[i] != a[i - 1]:
        alphabet.append(a[i])

point = 0
if alphabet[point] == 0:
    point += 1

if len(alphabet) == 1:
    cur = 2 * n
    if a[0] == 1:
        print(0, end=' ')
    else:
        print(cur, end=' ')

    for i in range(n - 1):
        print(cur, end=' ')
    print()
    exit()

curr = alphabet[point] - 1
mina = 0
for i in range(n):
    if a[i] == mina:
        if curr > mina and (point == 0 or curr != alphabet[point - 1]):
            print(curr, end=' ')
            curr -= 1
        else:
            point += 1
            while point < len(alphabet) and alphabet[point] - 1 == alphabet[point - 1]:
                point += 1

            if point == len(alphabet):
                curr = 2 * n
            else:
                curr = alphabet[point] - 1
            print(curr, end=' ')
            curr -= 1
    else:
        print(mina, end=' ')
        mina = a[i]