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.
ll = lambda: list(map(int, input().split()))
s = lambda: input()
v = lambda: map(int, input().split())
ii = lambda: int(input())
m=1000000007
n=ii()
a=ll()
m=[0]*(1000001)
c=[0]*(1000001)
for i in a:
    c[i]+=1
p=0
q=[]
ans=[]
for i in range(n+1):
    if(c[i]==0):
        continue
    while(c[i]):
        if(q):
            c[i]-=1
            ans.append(q[0])
            m[q[0]]+=1
            q.pop(0)
        else:
            while(c[p] or m[p]):
                p+=1
            m[p]+=1
            ans.append(p)
            p += 1
            c[i]-=1
    q.append(i)
print(ans)

