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.
#! /usr/bin/env python3
# -*- coding: utf-8 -*-
# vim:fenc=tf-8
#

"""
"""

from collections import defaultdict
from collections import Counter

import sys
input = sys.stdin.readline

def solve(n, a):
    m = max(a)
    sa = set(a)
    fill = [i for i in range(m,-1,-1) if i not in sa]
    ans = []
    a0 = 0
    for aa in a:
        if aa == a0:
            if fill:
                ans.append(fill.pop())
            else:
                ans.append(m+1)
        if aa > a0:
            if fill and fill[-1] == a0:
                fill.pop()
            if fill and fill[-1] < aa:
                print(-1)
                return 0
            else:
                ans.append(a0)
                a0 = aa
    print(" ".join(map(str,ans)))






def main():
    n = int(input())
    a = list(map(int,input().split()))
    solve(n, a)


if __name__ == "__main__":
    main()