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, math

input = lambda: sys.stdin.readline().rstrip()

n = int(input())
a = list(map(int, input().split()))
b = []
a = a[::-1]
b.append(a[0] - 1)
for i in range(1, n):
    b.append(a[i] - 1)
for i in range(n):
    if b[i] <= -1:
        b[i] = 1000002
b = b[::-1]
# print(*b)
ma = -1
for i in b:
    if i != 1000002 and i > ma:
        ma = i + 1
p = []
for i in range(ma):
    if i not in b:
        p.append(i)
if 0 in a:
    p[0] = 1000000
cur = 0
l = len(p)
for i in range(n):
    if b[i] == 1000002:
        b[i] = p[cur % l]
        cur += 1
print(*b)
