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
n=int(input());mark={};mark1={}
a=list(map(int,input().split()));t=0;Ans=[];p=deque()
for i in a:mark[i]=1
for i in a:
    if i==t:
        if len(p)>0:
            if i==p[0]:
                while mark.get(t,None)!=None:
                    if mark1.get(t,None)==None:
                        p.append(t);mark1[t]=1
                    t+=1    
                Ans.append(t);t+=1
            else:    
                q=p.popleft()
                Ans.append(q)
        else:        
            while mark.get(t,None)!=None:
                if mark1.get(t,None)==None:
                    p.append(t);mark1[t]=1
                t+=1    
            Ans.append(t);t+=1
    else:
        if len(p)>0:
            if i==p[0]:
                while mark.get(t,None)!=None:
                    if mark1.get(t,None)==None:
                        p.append(t);mark1[t]=1
                    t+=1    
                Ans.append(t);t+=1
            else:    
                q=p.popleft()
                Ans.append(q)
        else:
            Ans.append(t);t+=1
print(*Ans)            
    