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.
# https://codeforces.com/contest/1364/problem/C
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
# do magic here
n = int(input())
arr = [int(x) for x in input().split()]
if arr[0] > 1:
    print(-1)
    sys.exit()
last = arr[-1]
required = []
used = set()
i = n-1
MAX = int(1e6) - 1
ans = [0 for x in range(n)]
while i > 0:
    if arr[i] > i+1:
        print(-1)
        sys.exit()
    if arr[i-1] != arr[i]:
        ans[i] = arr[i-1]
        for x in range(arr[i-1]+1, arr[i]):
            required.append(x)
    else:
        if len(required):
            ans[i] = required.pop()
        else:
            ans[i] = MAX
    i -= 1
if arr[0] == 0:
    if len(required):
        ans[0] = required.pop()
    else:
        ans[0] = MAX
for item in ans:
    print(item, end=" ")
