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 = list(map(int, input().split(' ')))
temp = {}
counter = 0
ans = []
flag = False
for i in range(n):
    temp[a[i]] = True

if a[0] == 0:
    ans.append(1)
    counter = 2
elif a[0] == 1:
    ans.append(0)
    counter = 1
else:
    print(-1)
    flag = True

for i in range(1, n):
    if a[i] > i+1:
        print(-1)
        flag = True
    if a[i] != a[i-1]:
        ans.append(a[i-1])
    else:
        if not counter in temp:
            ans.append(counter)
            counter += 1
        else:
            while counter in temp:
                counter += 1
            ans.append(counter)
            counter += 1
    
if not flag:
    print(*ans)
        

    



    
    
    

    
        

    
