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.
from sys import stdin

def intJoin(arr):
  return " ".join([str(i) for i in arr])

def answer(arr):
  arr = [arr[0]-1] + arr
  a = []
  b = []
  for i in range(1, len(arr)):
    if arr[i] != arr[i-1]:
      a.append(arr[i-1])
      b.append(arr[i-1])
    else:
      b.append(-1)
  if arr[-1] - a[-1] > 1:
    a.append(arr[-1])
  c = []
  for i in range(1, len(a)):
    if (a[i] - a[i-1]) > 1:
      for j in range(a[i-1]+1, a[i]):
        c.append(j)
        
  c = list(reversed(c))
  for i in range(0, len(b)):
    if b[i] == -1:
      if len(c) == 0:
        b[i] = arr[-1]+1
      else:
        b[i] = c.pop()
  return intJoin(b)
  
  
n = int(input())
arr = list(map(int, stdin.readline().split()))
print(answer(arr))