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 = input()
arr = map(int,raw_input().split())

notUsed = set(range(N+1))

possible = True
lastOcc = {}
for n in range(N) :
  if arr[n] > n+1 : 
    possible = False
    break
  lastOcc[arr[n]] = n
  notUsed.discard(arr[n])

if possible :

  extra = list(notUsed)
  extra.sort()

  ans = [None] * (N+1)
  for k, v in lastOcc.items() : ans[v+1] = k

  i = 0
  for n in range(N) :
    if ans[n] is None : 
      ans[n] = extra[i]
      i += 1
  
  if ans[0] > 1 : print -1
  else : print ' '.join(str(x) for x in ans[:-1])

else : print -1