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,x = (int(i) for i in input().split())
n = int(input())
a = [int(i) for i in input().split()]

d = False
for i in range(n):
  if a[i]>i+1 or (i<n-1 and a[i+1]<a[i]):
    print(-1)
    d = True
    break

if not d:
  out = [-1]*n
  #if a[0] == 0:
  #  out[0] = '1'
  #else:
  #  out[0] = '0'
  for i in range(1,n):
    if a[i]!=a[i-1]:
      out[i] = str(a[i-1])
    else:
      out[i] = -1
  
  next_ = n

  seen = {a[-1]}
  for i in range(n-1,-1,-1):
    seen.add(a[i])
    if out[i] == -1:
      while next_ in seen:
        next_-=1
      out[i] = str(next_)
      next_-=1
  
  print(' '.join(out))


