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

t = 1
for _ in range(t):
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().split()))
    flg = True
    res = list()
    places = list()
    if a[0] not in (0, 1):
        flg = False
    p = a[0]
    cur = 0
    if p:
        res.append(0)
        cur += 1
    for i in range(1, n):
        if a[i] > i+1:
            flg = False
            break
        if a[i] != p:
            for j in range(p+1, a[i]):
                if cur >= i:
                    if not places:
                        flg = False
                        break
                    x = places.pop()
                    res[x] = j
                else:
                    res.append(j)
                    cur += 1
            while cur < i:
                res.append(1000000)
                places.append(cur)
                cur += 1
            res.append(p)
            cur += 1
        p = a[i]
    if flg:
        print(" ".join([str(x) for x in res]))
    else:
        print(-1)