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,stdout
import math 
from collections import Counter,deque
L=lambda:list(map(int, stdin.readline().strip().split()))
M=lambda:map(int, stdin.readline().strip().split())
I=lambda:int(stdin.readline().strip())
IN=lambda:stdin.readline().strip()
C=lambda:stdin.readline().strip().split()
mod=1000000007
#Keymax = max(Tv, key=Tv.get)
def s(a):print(" ".join(list(map(str,a))))
#______________________-------------------------------_____________________#
#I_am_pavan
def solve():
    n=I()
    a=L()
    ans=[-1]*n
    b=[0]*(n+1)
    if a[0]==1:
        ans[0]=0
        b[0]=-1
    for i in range(1,n):
        if a[i]>a[i-1]:
            ans[i]=a[i-1]
            b[a[i-1]]=-1
    j=0
    b[a[-1]]=-1
    #print(ans,b)
    for i in range(n):
        if ans[i]==-1:
            while b[j]==-1:
                j+=1
            ans[i]=j
            j+=1
    print(*ans)
for i in range(1):
    solve()
