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()))

def mex(n, a):
    if a[0] > 1:
        print(-1)
        return
    unuse = list(range(2*10**5+3,-1,-1))
    used = set(a)
    unused = [t for t in unuse if t not in used]
    ret = []
    i = 0
    while i < len(a):
        if i-1 >= 0 and a[i] != a[i-1]:
            ret.append(a[i-1])
            if unused[-1] <= a[i]:
                print(-1)
                return
        else:
            ret.append(unused.pop())
        i += 1
    print(' '.join(map(str, ret)))

mex(n, a)
