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
    used = [False]*a[j]
    for i in range(n):
        if sol[i] != -1:
            used[sol[i]] = True
    i = len(used)-1
    # print(used)
    while j >= 0:
        if sol[j] != -1:
            used[sol[j]] = True
        else:
            while i >= 0 and used[i]:
                i -= 1
            if i >= 0:
                sol[j] = i
                used[i] = True
            else:
                sol[j] = mx
        # print(used)
        j -= 1
    return sol
    # return sol if all(x != -1 for x in sol) else []

# 0 0 1 1 4 4 4 4 4 4
# 5 2 0 3 1 5 5 5 5 5

# x x x x

n = int(input())
a = list(map(int, input().split()))
s = solve(n, a)
print(' '.join(map(str, s)) if s else -1)