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())
list1=list(map(int,input().strip().split()))

if list1[0]>=2:
    print(-1)

else:


    free=[]
    curr=0

    b=1
    ans=[]
    for i in range(n):
        ans.append(-1)
        

    if list1[0]==0:
        free.append(0)
    elif list1[0]==1:
        ans[0]=0
        
        
    for i in range(1,n):
        if list1[i]==list1[i-1]:
            free.append(i)
        else:
            ans[i]=list1[i-1]
            
            temp1=len(free)-curr
            temp2=list1[i]-list1[i-1]-1
            
            if temp2>temp2:
                b=0
                break
            
            for i in range(list1[i-1]+1,list1[i]):
                ans[free[curr]]=i
                curr+=1

    if b==0:
        print(-1)
        
    else:
        maximum=list1[n-1]+1
        for i in range(curr,len(free)):
            ans[free[i]]=maximum

        for val in ans:
            print(val,end=" ")
        print()
        
        
    
    
        
        

