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())
l=list(map(int,input().split()))
r=l[-1]
ans=[0 for i in range(n)]
d=set()
for i in range(r):
   d.add(i)

p=list(d.difference(l))

ans[0]=1-l[0]
a=False
b=False
if l[0]==0:
    if 1 in p:
        p.remove(1)
    a=True
if l[0]==1:
    if 0 in p:
        p.remove(0)
    b=True
for i in range(1,n):
    if l[i]-i>1:
        print(-1)
        exit()
    if l[i]==0:
        a=True
    if l[i]==1:
        b=True
    if l[i]!=l[i-1]:
        ans[i]=l[i-1]
    else:
        ans[i]=ans[i-1]
q={}
p=p[::-1]

for i in range(n):
    if ans[i] in q:
        try:
            ans[i]=p.pop()
        except:
            ans[i]=r+1
    else:
        q[ans[i]]=1
if a and b:
    print(-1)
    exit()
print(*ans)
     
        
