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())
s=input().split()
flag=0
for j in range(n):
    s[j]=int(s[j])
    if s[j]>j+1:
        flag=1

if flag==1:
    print(-1)
else:
    if s[0]==1:
        for j in range (n-1):
            if s[j+1]-s[j]>1:
                flag=1
                break
        if flag==1:
            print(-1)
        else:
            for j in range (n):
                print(s[j]-1,end="")
    else:
        for j in range (n):
            if s[j]>=3:
                rr=j
            else:
                rr=0
        if rr:
            for j in range (rr,n-1):
                if s[j+1]-s[j]>1:
                    flag=1
                    break
            for j in range (n):
                if s[j]==1:
                    flag=1
        if flag==1:
            print(-1)
        else:
            for j in range (n):
                if s[j]==0:
                    print("1 ",end="")
                elif s[j]==2:
                    print("0 ",end="")
                else:
                    print(s[j]-1,end="")
        
    