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 = [int(j) for j in input().split(" ")]
b = [-1] * n

if a[0] > 1:
    print(-1)
else:

    if a[0] == 0:
        mex = 0
        count = 0
    else:
        mex = 1
        count = 1
        b[0] = 0

    result = 1
    for i in range(1, n):
        if a[i] != mex:
            b[i] = mex
            mex += 1
            while mex < a[i]:
                b[count] = mex
                mex += 1
                while b[count] != -1:
                    count += 1
                    if count > i:
                        if mex == a[i]:
                            break
                        else:
                            mex = a[i]
                            result = 0
                            break
            if result == 0:
                break

    if result == 1:
        for i in range(n):
            if b[i] == -1:
                b[i] = mex + 1
        for i in range(n - 1):
            print(b[i], end=' ')
        print(b[-1])
    else:
        print(-1)





