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 deque
import sys
n=int(input())
l=[-1 for i in range(n)]
a=[int(i) for i in input().split()]
maxx=max(a)
pos=deque()
flag=1
for i in range(n):
    if a[i]>i+1:
        flag=0
if a[0]==1:
    l[0]=0
else:
    pos.append(0)
for i in range(1,n):
    if a[i]==a[i-1]:
        pos.append(i)
        continue
    else:
        l[i]=a[i-1]
        for j in range(a[i-1]+1,a[i]):
            if len(pos)==0:
                flag=0
                break
            x=pos.popleft()
            l[x]=j
if flag:
    for i in pos:
        l[i]=maxx+1
    for i in l:
        print(i,end=" ")
    print
else:
    print(-1)
            
            
