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.readline


n = int(input())
a = list(map(int, input().split()))
ans = []
d = dict()
for i in range(n):
    if a[i] > i+1:
        print(-1)
        exit()

for i in range(10**5 + 20):
    d[i] = 0
for i in a:
    d[i] += 1

m = 10**6
for i in range(n):
    if d[i] == 0:
        m = i
        break
back_m = m
for i in range(n):
    ans.append(m)
    d[a[i]] -= 1
    if d[a[i]] == 0:
        m = a[i]
    else:
        m = back_m
        for j in range(m+1, n):
            if d[j] == 0:
                m = j
        back_m = m
print(*ans)
