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):
    sol = [-1]*n
    mx = a[-1]+1
    if a[0] == 1:
        sol[0] = 0
    for i in range(1, n):
        if a[i] != a[i-1]:
            sol[i] = a[i-1]
    j = n-1
    while sol[j] == -1:
        sol[j] = mx
        j -= 1
    if j > -1:
        used = [False]*a[j]
        for i in range(n):
            if sol[i] != -1 and sol[i] < len(used):
                used[sol[i]] = True
        i = len(used)-1
        while j >= 0:
            if sol[j] != -1:
                used[sol[j]] = True
            else:
                while i >= 0 and used[i]:
                    i -= 1
                if i >= 0 and i > a[j]:
                    sol[j] = i
                    used[i] = True
                else:
                    sol[j] = mx
            j -= 1
    return sol
 
n = int(input())
a = list(map(int, input().split()))
s = solve(n, a)
print(' '.join(map(str, s)) if s else -1)