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 collections import Counter
from collections import defaultdict
from collections import deque
import math
import heapq
import sys
import io, os

input = sys.stdin.readline
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import *

rs = lambda: input().strip()
ri = lambda: int(input())
rl = lambda: list(map(int, input().strip().split()))
rls = lambda: list(map(str, input().split()))

n = int(input().strip())
a = rl()
s = list(set(a))
s.sort()
d = defaultdict(lambda: -1)
ans = []
prev = 0
for i in s:
    d[i] = 1
f = 1
prev=-1
for i in range(0, n):
    if ( i!=0 and a[i] != a[i - 1]):
        ans.append(a[i - 1])
        continue
    while (1):
        prev = prev + 1
        # print(i,prev)
        if (d[prev] == 1):
            continue
        else:
            break
    #  print(i,prev)
    ans.append(prev)
    if (a[i] - prev > 1):
        f = 0
        break
if (f):
    mx = 0
    for i in range(0, n):
        mx = max(ans[i], mx)
        if (a[i] - mx > 1):
            print(-1)
            exit()
    print(*ans)
else:
    print(-1)
