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.
def f(a):
    tp=0
    cmax=max(a)
    if cmax==0:
        return -1
    st=set(a)
    included=[0]*(len(a)+1)
    ans=[]
    ahead=[0]*(len(a)+1)
    for i in st:
        ahead[i]=-1

    for val in a:
        flag = False
        if val==0:
            ans.append(1)
            continue
        for i in range(tp,len(ahead)):
            if included[i]==0 and i!=val and ahead[i]!=-1:
                ans.append(i)
                included[i]=1
                flag=True
                tp=val
                ahead[val]=0
                break
        if flag==False:
            ans.append(tp)
    return ans


a=input()
lst=list(map(int,input().strip().split()))
print(*f(lst))