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.
n=int(input())
l=input().split()
li=[int(i) for i in l]
lo=[0 for i in range(n)]
set=0
done=[0 for i in range(n)]
hashi=dict()
for i in li:
    hashi[i]=1
for i in range(n):
    if(li[i]!=0 and i==0):
        lo[i]=0
        done[i]=1
        hashi[0]=1
        continue
    if(li[i]!=li[i-1] and i!=0):
        lo[i]=li[i-1]
        done[i]=1
curr=0
for i in range(n):
    if(done[i]!=0):
        continue
    while(curr in hashi):
        curr+=1

    lo[i]=curr
    curr+=1
for i in lo:
    print(i,end=" ")
