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())
a=[int(f) for f in input().split()]
m=set(a)
b=[]
p=set(f for f in range(0,a[n-1]+1))
p=p-m
p=list(p)
##print(p)
g=0
l=0
cur=0
fl=True
for i in range(n):
    if i==0:
        if a[i]==0:
            b.append(p[cur])
            cur+=1
        else:
            ##a[i]==1
            b.append(0)
            if p[cur]==0:
                cur+=1
        if cur==len(p):
            fl=False
    else:
        if a[i]==a[i-1]:
            if fl:
                b.append(p[cur])
                cur+=1
            else:
                b.append(a[n-1]+10)
        else:
            b.append(a[i-1])
            

        if cur==len(p):
            fl=False
        
print(b)
    
