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 sys import stdin, stdout  

n=int(stdin.readline() )
#n,x=(map(int, stdin.readline().strip().split()))
arr=list(map(int, stdin.readline() .strip().split()))
#s=stdin.readline()

#stdout.write(str(cp))
#stdout.write("\n")

if(sorted(arr)!=arr):
    print(-1)
else:
    flg=0
    for i in range(n):
        if(arr[i]-i>1):
            flg=1
            break
    if(flg):
        print(-1)
    else:
        raj=dict()
        for i in range(n):
            raj[arr[i]]=1
        x=-1
        ans=[]

        for i in range(n):
            if(i==0):
                 x=x+1
                 while(x in raj):
                     x+=1
                 ans.append(x)
            else:
                if(arr[i-1]!=arr[i]):
                    ans.append(arr[i-1])
                else:
                    x=x+1
                    while(x in raj):
                        x+=1
                    ans.append(x)
        print(*ans)
