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 sys import stdin, stdout, setrecursionlimit
for _ in range(1):
    n = int(input())
    a = [int(i) for i in input().split()]
    f = True
    ans = [-1] * n
    ok = [False] * (n + 1)
    for i in range(n):
        if a[i] > i + 1 or a[i] > a[min(i + 1, n - 1)]:
            f = False
            break
        ok[a[i]] = True
    if not f:
        print(-1)
        continue
    nums = []
    curi = 0
    for i in range(n + 1):
        if not ok[i]:
            nums.append(i)
    for i in range(n - 1):
        if a[i] != a[i + 1]:
            ans[i + 1] = a[i]
    for i in range(n):
        if ans[i] == -1:
            ans[i] = nums[curi]
            curi += 1

    print(*ans)

