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, = map(int, input().split())

    for i, a in enumerate(A):
        if i+1 < a:
            print(-1)
            return
    else:
        Us = set(range(N+1)) - set(A)
        U = sorted(Us)
        lU = len(U)

        ui = 0
        ans = []
        pre = A[0]
        for i, a in enumerate(A):
            if pre != a:
                ans.append(pre)
            else:
                if ui >= lU:
                    print(-1)
                    return
                ans.append(U[ui])
                ui += 1
            pre = a

        print(*ans)

if __name__ == '__main__':
    main()
