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()))
left=[]
i=0
for x in range(a[n-1]+1):
    if a[i]>x:
        left.append(x)
    elif a[i]<=x:
        while i<n and a[i]<=x:
            i+=1
left.append(a[n-1]+1)
if a[0]!=0 and a[0]!=1:
    print('-1')
else:
    b=[left[0]]
    j=1
    k=0
    while j<n:
        if a[j]!=a[j-1]:
            b.append(a[j-1])
            j+=1
        else:
            if left[k+1]<a[j+1]:
                b.append(left[k+1])
                k+=1
                j+=1
            else:
                b.append(b[j-1])
                j+=1
    for i in range(n):
        print(b[i],end=' ')
