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 sys, heapq

max_int = 1000000001  # 10^9+1
min_int = -max_int

n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))

d = {}
maxa = max(a)
to_add = set(range(0, maxa + 2))

for num, one in enumerate(a):
    if one - num >= 2:
        print(-1)
        exit()
    d[one] = num
    to_add.discard(one)

h = list(to_add)
del to_add
heapq.heapify(h)

out = []
last = -1
for num, elem in enumerate(a):
    if h:
        last = heapq.heappop(h)
    out.append(last)

    if num == d[elem]:
        heapq.heappush(h, elem)

print(' '.join(map(str, out)))
