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

# import math
# from collections import deque
# import heapq
# from math import inf
# from math import gcd

# print(help(deque))
# 26
pprint = lambda s: print(' '.join(map(str, s)))
input = lambda: sys.stdin.readline().strip()
ipnut = input

# n,x = map(int,input().split())
n = int(input())
a = [0] + list(map(int, input().split()))
ans = [-1] * n
used = set(a)
for i in range(n):
    if a[i] != a[i + 1]:
        ans[i] = a[i]
    if a[i+1]>i+1:
        print(-1)
        exit(0)
# print(ans)
curr = 0
for i in range(n):
    if ans[i]==-1:
        curr+=1
        while curr in used:
            curr+=1
        ans[i]=curr
pprint(ans)