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.
#!/usr/bin/env python3
import io
import os

input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline


def get_str():
    return input().decode().strip()


def rint():
    return map(int, input().split())


def oint():
    return int(input())


n = oint()
a = list(rint())

ans = [-1] * n
used = set()

for i in range(1, n):
    if a[i - 1] == a[i]:
        continue
    ans[i] = a[i - 1]
    used.add(ans[i])

used.add(a[-1])
cur = 0
for i in range(n):
    if ans[i] != -1:
        continue
    for j in range(cur, 10 ** 9, 1):
        if j not in used:
            ans[i] = j
            cur = j + 1
            break
print(*ans)
