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())
    lst = list(map(int, input().split()))
    reserve = set()
    if lst[0] > 1:
        print(-1)
        return
    for i in range(0, n):
        if lst[i] > i + 1:
            print(-1)
            return
        reserve.add(lst[i])
    can_use = []
    cur = lst[0]
    for i in range(0, n + 1):
        if not i in reserve:
            can_use.append(i)
    can_use.reverse()
    used = []
    for i in range(0, n):
        if lst[i] == cur:
            if len(can_use) == 0:
                print(-1)
                return
            used.append(can_use.pop())
        else:
            if len(can_use) != 0 and can_use[-1] < cur:
                print(-1)
                return
            used.append(cur)
            cur = lst[i]
    line = ''
    for i in used:
        line += str(i) + ' '
    print(line)


if __name__ == '__main__':
    t = 1
    for i in range(t):
        main()
