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
# sys.stdin = open('input.txt', 'r') 
# sys.stdout = open('output.txt', 'w')
 
import math
import collections
from sys import stdin,stdout,setrecursionlimit
import bisect as bs
setrecursionlimit(2**20)
M = 10**9+7
 
# T = int(stdin.readline())
T = 1      

for _ in range(T):
    n = int(stdin.readline())
    # a,b = list(map(int,stdin.readline().split()))
    a = list(map(int,stdin.readline().split()))
    # b = list(map(int,stdin.readline().split()))
    # q = list(map(int,stdin.readline().split()))
    # b = list(map(int,stdin.readline().split()))
    # s = stdin.readline().strip('\n')
    d = [-1]*(n)
    done = [False]*(n+1)
    s = 0
    for i in range(n-1):
        if(a[i] == a[i+1]):
            continue
        d[i+1] = a[i]
        done[a[i]] = True
        s = i+1
    d[0] = -1 if a[0] == 0 else 0
    if(a[0] == 1):
        done[0] = True
    done[a[-1]] = True
    fin = []
    for i in range(n,-1,-1):
        if(not done[i]):
            fin.append(i)
    for i in range(n):
        if(d[i] == -1):
            d[i] = fin.pop()  
    for h in d:
        print(h, end =' ')
    print('')