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 io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def f():
    n = int(input())
    A = [int(s) for s in input().split()]
    if A[-1]==0:
        # A all 0
        B = [1]*n
        print(' '.join(str(num) for num in B))
        return
    for i in range(n):
        if A[i] > i+1:
            print(-1)
            return

    l = A[-1]
    visited = [0]*l
    B = [-1]*n

    if A[0] >= 2:
        print(-1)
        return
    if A[0] == 1:
        B[0] = 0
        visited[0] = 1
    for i in range(1,n):
        if A[i] != A[i-1]:
            B[i] = A[i-1]
            visited[A[i-1]] = 1
    # all ele in B should be seen first time
    i = 0
    for j in range(l):
        if visited[j]:
            continue
        while B[i] != -1:
            i += 1
            if i >= n:
                print(-1)
                return
        B[i] = j
    for i in range(n):
        if B[i] == -1:
            B[i] = l+1

    print(' '.join(str(num) for num in B))


f()