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.
a = int(input())
b = [int(x) for x in input().split()]
impo = 0
uniq =[]
notyet = []
if(b[0]!=0):
    notyet.append([0,b[0]])
k = [-1]*a
for i in range(a):
    if(i+1<b[i]):
        impo = 1
        break
    if(b[i]>b[i-1]):
        k[i] = b[i-1]
        uniq.append(k[i])
        if(b[i]-b[i-1]!=1): 
            notyet.append([b[i-1]+1,b[i]])
#print(notyet)
notyeti = 0
lack = 0
if(impo==1):
    print(-1)
else:
    for i in range(a):
        if (k[i]==-1):
            if(notyeti<len(notyet)):
                k[i] = notyet[notyeti][0]
                notyet[notyeti][0]+=1
                if(notyet[notyeti][0]==notyet[notyeti][1]):
                    notyeti+=1
            else:
                k[i] = b[-1]+1
    for i in k:
        print(i,end=" ")
    print()
        
        
    