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.
import sys

input = sys.stdin.readline


def main():
    N = int(input())
    A = [int(x) for x in input().split()]

    seta = set(A)
    mex = 0

    ans = []
    for i, a in enumerate(A):
        if a == mex:
            if len(ans) == 0:
                for j in range(1, N + 1):
                    if j not in seta:
                        ans.append(j)
                        break
                else:
                    print(-1)
                    return
            else:
                for j in range(max(mex, ans[-1]) + 1, N + 1):
                    if j not in seta:
                        ans.append(j)
                        break
                else:
                    print(-1)
                    return
        else:
            ans.append(mex)
            mex = a

    print(*ans)


if __name__ == '__main__':
    main()
