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.
i_liczb = int(input())
a = [int(x) for x in input().split()]
a = [0] + a 
if a[i_liczb] > i_liczb:
    print(-1)
    quit()
b = [-1 for _ in range(i_liczb + 1)]
wykorzystane = [False for _ in range(100005)]
for i in range(1, i_liczb + 1):
    if a[i] !=a [i - 1]:
        b[i] = a[i - 1]
        wykorzystane[b[i]] = True
wykorzystane[a[i_liczb]] = True
x = 0
for i in range(1, i_liczb + 1):
    while wykorzystane[x]:
        x += 1
    if b[i] == -1:
        b[i] = x
        wykorzystane[x] = True
print(*b[1:])
