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.
from collections import Counter
n = input()
a = map(int,raw_input().split())
ans = []
d = Counter(a)
target = max(a)-1
if a[0]==0:
    ans.append(1)
    p = 1
else:
    ans.append(0)
    p = 0
for i in range(1,n):
    if a[i]!= a[i-1]:
        ans.append(a[i-1])
        p = a[i-1]
    else:
        
        q = p + 1
        
        while 1:
            if q==target+1:
                break
            if d[q]==0:
                break
            
            q+=1
        
        if q==target+1:
            ans.append(p)
        else:
            
            ans.append(q)
            p = q
for x in ans:
    print x,
print 
