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.
from collections import deque
from sys import stdin, stdout

int_in = lambda: int(stdin.readline())
arr_in = lambda: [int(x) for x in stdin.readline().split()]
mat_in = lambda rows: [arr_in() for y in range(rows)]
str_in = lambda: stdin.readline().strip()
out = lambda o: stdout.write("{}\n".format(o))
arr_out = lambda o: out(" ".join(map(str, o)))
bool_out = lambda o: out("YES" if o else "NO")
tests = lambda: range(1, int_in() + 1)
case_out = lambda i, o: out("Case #{}: {}".format(i, o))


def solve(n, a):
    result = [None] * n
    items_in_a = set(a)
    curr = 0
    for i in range(n):
        if i > 0 and a[i - 1] != a[i]:
            result[i] = a[i - 1]
        else:
            while curr in items_in_a:
                curr += 1
            result[i] = curr
            curr += 1
    return result


if __name__ == "__main__":
    n = int_in()
    a = arr_in()
    arr_out(solve(n, a))