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 get_not_elements(a:list):
    elements = set(a)
    max_element = a[-1] +1
    not_elements = []
    for i in range(0,max_element):
        if not i in elements:
            not_elements.append(i)
    not_elements.reverse()
    return not_elements

#Ahora pensar esta funcion...
def is_valid(a : list):

    for i in range(len(a)):
        if a[i]-1 > i:
            return False
    return True

def generate_array(a : list):
    
    stack_ = get_not_elements(a)
    b = [-1 for i in range(len(a))]
    val = a[0]
    for i in range(1,len(a)):
        if a[i] != val:
            b[i] = val
            val = a[i]

    if len(stack_)>0:
        val = stack_.pop()
    else:
        val = a[0]

    for i in range(len(a)):
        if b[i] == -1:
            b[i] = val

            if len(stack_)>0:
                val = stack_.pop()
    return b 


if __name__ == "__main__":
    size  =int(input())
    array = [int(i) for i in input().split()]


    if not is_valid(array):
        print(-1)
    else:
        b = generate_array(array)
        b = " ".join([str(i) for i in b])
        print(b)
    