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.
import bisect
n = int(input())
a = list(map(int, input().split()))
s = set()
b = [0]*n
m = 0
s = set()
for i in range(n-1, -1, -1):
    if i == 0:
        if a[i] == 1:
            b[i] = 0
        else:
            if len(s) > 0:
                b[i] = s.pop()
            else:
                b[i] = n + 1
    else:
        if a[i] != a[i-1]:
            b[i] = a[i-1]
            for x in range(a[i-1]+1, a[i]):
                s.add(x)
        else:
            if len(s) > 0:
                b[i] = s.pop()
            else:
                b[i] = n + 1

# T = set()


# def checker(A, B):
#     s = set()
#     n = len(A)
#     print(*A)
#     print(*B)
#     for i in range(n):
#         s.add(b[i])
#         k = n+1
#         for j in range(n+1):
#             if j in s:
#                 continue
#             else:
#                 k = j
#                 break
#         if k != a[i]:
#             return(False)
#     return(True)


print(*b)


# print(checker(a, b))
