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, a):

    fail = [-1]
    b = [None] * n
    s = {}
    t = []
    m = max( a )

    for i in range(n):
        if a[i] > i+1:
            return fail

        if i==0:
            if a[i]==1:
                b[i] = 0
                s[0] = 1
            continue

        if a[i] > a[i-1]:
            b[i] = a[i-1]
            s[ a[i-1] ] = 1


    for i in range(n-1, -1, -1):
        if b[i] is not None:
            for j in range(a[i]-1, -1, -1):
                if j not in s:
                    t.append(j)
        else:
            if len(t) > 0:
                b[i] = t.pop()
            else:
                b[i] = m



    return b



n = int(input().strip())

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

print(' '.join(map(str, solve(n, a))))
