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
import heapq
import collections
def inputnum():
    return(int(input()))
def inputnums():
    return(map(int,input().split()))
def inputlist():
    return(list(map(int,input().split())))
def inputstring():
    return([x for x in input()])
def inputmatrixchar(rows):
    arr2d = [[j for j in input().strip()] for i in range(rows)] 
    return arr2d
def inputmatrixint(rows):
    arr2d = []
    for _ in range(rows):
        arr2d.append([int(i) for i in input().split()])
    return arr2d

n = inputnum()
a = inputlist()
b = [-1]*(n)
c = [False]*(1000000)
c[a[0]] = True
for i in range(1, n):
    c[a[i]] = True
    if a[i] > a[i-1]:
        b[i] = a[i-1]
m = 0
for i in range(n):
    while c[m] == True:
        m += 1
    if b[i] == -1:
        b[i] = m 
        c[m] = True
print(*b)