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.

import math

n = int(input())
array = list(map(int,input().split()))
dict_num = {}
ans = 0
for i in range(n):
    if array[i]>i+1 or (i+1<n and array[i]>array[i+1]): 
        ans = -1
        break
    dict_num[array[i]]=0

# sortedArray = sorted(list(set(array)))

if ans==-1:print(-1)
else:
    stored = []
    not_stored = []
    for i in range(n,-1,-1):
        if i in dict_num: stored.append(i)
        else:not_stored.append(i)
    ans = []
    for i in range(n):
        # print(stored)
        # print(not_stored)
        # print(ans)
        if ans==[]: 
            ans.append(not_stored[-1])
            not_stored.pop()
        else:
            new = 0
            if dict_num[stored[-1]]==1:
                new = stored[-1]
                stored.pop()
            else:
                new  = not_stored.pop()

            ans.append(new)

        if i+1<n:
            # print(i,array[i])
            if array[i]!=array[i+1]:
                # print("changed",array[i])
                dict_num[array[i]] = 1
    print(*ans)