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.
n = int(input())
a = list(map(int, input().split()))

b = [-1 for i in range(n)]
flag = 1
c = a[0]
r = {}
for i in range(1, n):
    if a[i] != c:
        b[i] = c
        r[c] = 1
        if i < c:
            flag = 0
            break
        c = a[i]
# print(c)
# b[n - 1] = c
# r[c] = 1
if n < c:
    flag = 0

avail = []
la = 0
for i in range(a[n - 1]):
    if r.get(i, 0) == 0:
        avail.append(i)
        la += 1
la += 1
avail.append(a[n - 1] + 1)
if flag == 0:
    print(-1)
else:
    j = 0
    for i in range(n):
        if b[i] == -1:
            b[i] = avail[j]
            if j < la - 1:
                j += 1
    for i in b:
        print(i, end=' ')