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.
# OBSERVE

# if i>a[i], -1. (i starts from 1)
# if a[i] != a[i-1], b[i] = a[i-1]

import sys
from sys import setrecursionlimit
setrecursionlimit(10 **5)

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

readline = sys.stdin.readline

n = int(readline())
a = list(map(int, readline().split()))
b = [-1] *n
not_visited = []
visited = [-1] * 10**6

flag = True
for i,j in enumerate(a):
    if j>i+1:
        flag = False
        break
    visited[j] = 1

if not flag:
    print(-1)
    sys.exit(0)

for i in range(1, n):
    if a[i] != a[i-1]:
        b[i] = a[i-1]

for i,j in enumerate(visited):
    if j == -1:
        not_visited.append(i)

k=0
for i,j in enumerate(b):
    if j==-1:
        b[i] = not_visited[k]
        k+=1

print(*b)




