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 main():
    n = int(input())
    a = list(map(int, input().split(' ')))

    if a[0] > 1:
        print(-1)
        return

    b = [-1] * n
    f = []
    fi = 0

    if a[0] == 1:
        b[0] = 0
    else:
        f.append(0)

    for i in range(1, n):
        if a[i - 1] == a[i]:
            f.append(i)
        else:
            b[i] = a[i - 1]
            for x in range(a[i - 1] + 1, a[i]):
                if len(f)==fi:
                    print(-1)
                    return
                b[f[fi]] = x
                fi+=1

    for j in f[fi:]:
        if j == 0:
            b[j] = 1
        else:
            b[j] = b[0]

    print(*b, sep=' ')


# for t in range(int(input())):
main()
