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, stdout
import math
import bisect

def main():
    n = int(stdin.readline())
    l = list(map(int, stdin.readline().split()))

    for i, x in enumerate(l):
        if x > i+1:
            print(-1)
            return 
    missing = list(set(list(range(n+1)))-set(l))[::-1]
    l = [0]+l
    ans = []

    for i in range(1, n+1):
        if l[i] == l[i-1]:
            ans.append(missing.pop())
        else:
            ans.append(l[i-1])
    stdout.write(" ".join(map(str, ans))+"\n")
main()
