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.
import sys

read = lambda: sys.stdin.readline()

n = int(read())
arr = list(map(int, read().split()))
result = [-1 for i in range(n)]
for i in range(n - 1, 0, -1):
    if(arr[i] > arr[i - 1]):
        result[i] = arr[i - 1]
if(arr[0] == 1):
    result[0] = 0
haves = set([])
free = 0
while(free < n and result[free] != -1):
    free += 1
impossible = False
for i in range(1, n):
    if(result[i] == -1):
        continue
    haves.add(result[i])
    for j in range(arr[i - 1], arr[i]):
        if(not j in haves):
            if(free > i):
                impossible = True
            result[free] = j
            haves.add(j)
            free += 1
            while(free < n and result[free] != -1):
                free += 1
for i in range(n):
    if(result[i] == -1):
        result[i] = 100001
if(impossible):
    print(-1)
else:
    for i in range(n - 1):
        print(result[i], end = " ")
    print(result[-1])
            