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 copy import *
from math import *
from sys import *

# t=int(input())
# for i in range(t):
n=int(input())
a=list(map(int,input().split()))

b=[]
if a[0]>1:
    print(-1)
    exit()

for i in range(len(a)-1):
    if a[i]==0 and a[i+1]==1:
        print(-1)
        exit()
    if a[i]>0:
        break


i=0
daddy=True
while i<len(a)-1:
    if a[i]>i+1:
        print(-1)
        exit
   


    elif a[i+1]-a[i]==2:
        b.append(a[i+1]-1)
        b.append(a[i+1]-2)
        i+=1
    # daddy=False

    elif a[i]==0:
        b.append(1)

    elif a[i+1]-a[i]==1:
        b.append(a[i]-1)

    elif a[i+1]==a[i]:
        b.append(a[i]-1)
    
    if len(b)==len(a):
        daddy=False
    i+=1
    
if daddy==True:
    b.append(a[len(a)-1]-1)
print(*b)

   