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())
a = list(map(int,input().split()))
a1= sorted(a)
if a1!=a:
    print(-1)
else:
    m = max(a)-1
    mm=m+10
    ans=[-1]*n
    has=[0]*(n+10)
    for i in range(n-1,-1,-1):
        has[a[i]]=1
        if a[i]>i+1:
            print(-1)
            exit()
        else:
            if i>0:
                if a[i-1]!=a[i]:
                    ans[i]=a[i-1]
    c=0
#    print(ans)
    for i in range(n):
        if ans[i]!=-1:
            continue
        while has[c]:
            c+=1
        ans[i]=c
        has[c]=1    
    print(*ans)        





