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 collections import defaultdict, Counter
from math import sqrt, log10, log2, log, gcd, floor, factorial
from bisect import bisect_left, bisect_right
from itertools import combinations, combinations_with_replacement
import sys, io, os
input=lambda:sys.stdin.readline().encode()
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# print=sys.stdout.write
# sys.setrecursionlimit(10000)
mod = 10 ** 9 + 7;inf = float('inf')
def get_list(): return [int(i) for i in input().split()]
yn = lambda a: print("YES" if a else "NO")
ceil = lambda a, b: (a + b - 1) // b
t=1

for i in range(t):
    n=int(input())
    l=[0]+[int(i) for i in input().split()]
    ans=[0]*(n+1)
    stack=[]
    initial=0
    dict=defaultdict(int)
    for i in range(1,n+1):
        if l[i]>l[i-1]:
            ans[i]=(l[i-1])
            dict[l[i-1]]=1
            while initial<l[i]:
                if initial in dict:
                    initial+=1
                    continue

                ans[stack.pop()]=initial
                initial+=1
        else:
            stack.append(i)
            ans[i]=10**6

    print(*ans[1:])




