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()))
L = [-1 for _ in range(n)]
L[0] = (A[0] == 0)*1
F = [0 for i in range(n)]
flg = True
F[0] = 1
for i in range(1, n):
    if A[i-1] != A[i]:
        L[i] = A[i-1]
        F[i] = 1
prev = 0
pos = 0
for j in range(n):
    if A[j] != prev:
        cnt = 0
        while cnt < A[j] - prev - 1:
            if F[pos] == 0:
                F[pos] = 1
                cnt += 1
                L[pos] = prev + cnt
            pos += 1
            if pos == j and cnt < A[j] - prev - 1:
                flg = False
                break
        prev = A[j]
    if not flg:
        break
if flg:
    for i in range(n):
        if F[i] == 0:
            L[i] = L[i-1]
    for i in range(n):
        print(L[i], end = ' '*(i!=n-1))
else:
    print(-1)