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.
n = int(input())
aarr = list(map(int, input().split()))

tovisit = {t : 0 for t in range(aarr[-1] + 1)}

for k in aarr:
    tovisit[k] += 1

b = []
transit = []
i = 0
flag = True

for t in aarr:
    if t > i + 1:
        flag = False
        print(-1)
        break
    if transit:
        b.append(transit[0])
        del transit[0]
        tovisit[t] -= 1
        if tovisit[t] == 0 and t < i:
            transit.append(t)
        continue
    while i < aarr[-1] + 1 and tovisit[i]:
        i += 1
    
    b.append(i)
    i += 1
    tovisit[t] -= 1
    if tovisit[t] == 0 and t < i:
        transit.append(t)


if flag:
    print(*b)
