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 *
from heapq import *
input = stdin.readline

n = int(input())
a = [int(x) for x in input().split()]
b = list(range(1, n + 1))
chk = [0] * (n + 1)
flag = 0
q = []

for x in range(n):
    if a[x] > b[x]:
        flag = -1
    chk[a[x]] += 1

if flag == -1:
    print(-1)
    exit()

for x in range(n + 1):
    if chk[x] == 0:
        heappush(q, x)

t = heappop(q)
print(t, end = ' ')

for x in range(1, n):
    if a[x - 1] != a[x]:
        print(a[x - 1], end = ' ')
    else:
        t = heappop(q)
        print(t, end = ' ')