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=list(map(int,input().split()))
totalmax=max(l)
s=set(range(1,l[-1]))
flag=True
ans=[]
reduntspots=[]
prev=-1
for i in range(n):
    if l[i]>(i+1):
        flag=False
        print(-1)
        break
    if l[i]==1:
        ans.append(0)
        prev=1
    elif l[i]==0:
        ans.append(totalmax+1)
        prev=0
    if l[i]==l[i-1]:
        reduntspots.append(i)
    if l[i] in s:
        s.remove(l[i])
    
    if l[i]!=0 and l[i]!=1:
        if prev!=l[i]:
            ans.append(prev)
            prev=l[i]
        else:
            ans.append(ans[-1])

if flag:
    missingcnt=len(s)
    if missingcnt==0:
        print(*ans)
    elif (missingcnt > len(reduntspots)):
        print(-1)
    else:
        missinglist = list(sorted(s))
        for i in range(missingcnt):
            ans[reduntspots[i]]=missinglist[i]
        print(*ans)


    