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.



if __name__ == "__main__":
    n = int(input())
    s = list(map(int,input().split()))
    stack = list()
    for i in range(n,s[-1],-1):
        stack.append(i)
    for i in range(len(s)-1,0,-1):
        if(s[i]==s[i-1]):
            continue
        else:
            for j in range(s[i]-1,s[i-1],-1):
                stack.append(j)
    fin_list = list()
    if(s[0]==1):
        stack.append(0)                 
    for i in range(len(s)-1):
        if(s[i]!=s[i+1]):
            curr_val = stack.pop(-1)
            fin_list.append(curr_val)
            stack.append(s[i])
        else:
            curr_val = stack.pop(-1)
            fin_list.append(curr_val)
    curr_val = stack.pop(-1)
    fin_list.append(curr_val)
    print (*fin_list)
