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()))

A.insert(0, 0)
output = [-1]*(n+1)
flag = 0
last = 0
for i, a in enumerate(A):
    if a > i+1:
        flag = 1
        break
    if i > 0:
        if A[i] > A[i-1]:
            output[i] = A[i-1]
if flag:
    print(-1)
else:
    j = 1
    while j < n+1 and output[j] != -1:
        j += 1
    for i, o in enumerate(output):
        if o != -1:
            for k in range(o+1, A[i]):
                output[j] = k
                while j < n+1 and output[j] != -1:
                    j += 1
    for i in range(1, len(output)):
        o = output[i] if output[i] != -1 else A[n]+1
        print(o, end=' ')
    print()
