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 math
import sys
from itertools import permutations
input = sys.stdin.readline

n=int(input())
arr=list(map(int,input().split()))
if arr[-1]=="\n":
    arr.pop()

ans=[-1]*n
tot=[0]
for i in range(n-1):
    if arr[i]!=arr[i+1]:
        ans[i+1]=arr[i]
        tot.append(i+1)

if n==1:
    pass
else:
    if arr[-1]!=arr[-2]:
        tot.append(n-1)

cur=0
done=0
# print("ans",ans)
# print("tot",tot)

for i in range(n):
    if ans[i]==-1:
        while cur<len(tot) and done==arr[tot[cur]]:
            done+=1
            cur+=1
        if cur>len(tot):
            break
        ans[i]=done
        done+=1

        

print(*ans)