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 sys
input = sys.stdin.buffer.readline


n = int(input())
a = list(map(int, input().split()))
INF = 10 ** 6 - 1

ans = [INF] * n
l = 0
if a[0] == 1:
    ans[0] = 0

for i in range(n - 1):
    if i + 1 < a[i]:
        print(-1)
        exit()

    if a[i] < a[i + 1]:
        if i + 2 < a[i + 1]:
            print(-1)
            exit()
        ans[i + 1] = a[i]
        min_num = a[i] + 1
        max_num = a[i + 1] - 1
        while min_num <= max_num:
            if ans[l] == INF:
                ans[l] = min_num
                min_num += 1
            l += 1
print(*ans)