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=[int(x) for x in input().split()]
if(l[0]>1 or l[-1]>n):
    print(-1)
else:
    flag=0
    dup=0
    for i in range(n):
        if(l[i]-i>1):
            flag=1
            break
    if(flag==1):
        print(-1)
    else:
        s1=set(l)
        s2=set([int(x) for x in range(0,n+2)])
        s3=s2.difference(s1)
        l1=list(s3)
        l1.sort()
        ans=[]
        if(l[0]==0):
            ans.append(l1.pop(0))
        else:
            l1.pop(0)
            ans.append(0)
        prev=l[0]
        for i in range(1,n):
            if(l[i]==prev):
                ans.append(l1.pop(0))
            else:
                ans.append(prev)
                prev=l[i]
        print(*ans)

            