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()))
b = [n+1] * n
isCovered = [False] * a[n-1]
lastIsNotCovered = 0
for i in range(n-1):
    if a[i] != a[i+1]:
        b[i+1] = a[i]
        isCovered[a[i]] = True
for i in range(lastIsNotCovered,a[n-1]):
    if isCovered[i] == False:
        lastIsNotCovered = i
        break
    if i == a[n-1] - 1:
        allCovered = True
        break

allCovered = False
if a[n-1] == 0:
    allCovered = True
for i in range(n):
    if allCovered:
        break
    if b[i] != n+1:
        continue
    else:
        b[i] = lastIsNotCovered
        if lastIsNotCovered == a[n-1] - 1:
            allCovered = True
            break
        isCovered[lastIsNotCovered] = True
        for j in range(lastIsNotCovered,a[n-1]):
            if isCovered[j] == False:
                lastIsNotCovered = j
                break
            if j == a[n-1] - 1:
                allCovered = True
                break
        if allCovered:
            break

if allCovered:
    output = [str(i) for i in b]
    print(' '.join(output))
else:
    print(-1)