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:
    store = []
    num = [i for i in range(max(array)-1,-1,-1)]
    ans = []
    for i in range(n):
        new  = num[-1]
        if new == array[i]: 
            if len(num)>1: 
                ans.append(num[-2])
                t = num.pop()
                num.pop()
                num.append(t)
            else:
                ans.append(ans[-1])    
        else:
            ans.append(new)
            num.pop()
    # print(len(ans))
    print(*ans)

    
            


