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.
n=int(input())
ans=[-1]*n 
se=set()
lis=list(map(int,input().split()))
length=len(lis)
for i in range(length-1,0,-1):
    if lis[i-1]!=lis[i]:
        ans[i]=lis[i-1]
        se.add(lis[i-1])
if lis[0]>1:
    print(-1)
    exit()
i=0
se.add(lis[-1])
j=0
while(True):
    
    if j==length:
        break
    if i in se :
        i+=1
        continue
    if ans[j]!=-1:
        j+=1 
        continue
    ans[j]=i
    j+=1
    i+=1
for i in ans:
    print(i,end=" ")