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.
#!/usr/bin/env python3
import io
import os

input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

def get_str():
    return input().decode().strip()

def rint():
    return map(int, input().split())

def oint():
    return int(input())

n = oint()
a = list(rint())

ans = [-1] * n

s = set()
if a[0] == 1:
    ans[0] = 0
    s.add(0)

for i in range(1, n):
    if a[i-1] == a[i]:
        continue
    ans[i] = a[i-1]
    s.add(ans[i])

f = a[-1] -1
for i in range(n):
    if ans[i] != -1:
        continue
    if f == 1 or f == 0:
        ans[i] = f
        continue
    for j in range(f, -1, -1):
        if not (j in s):
            ans[i] = j
            f = j-1
            break
    else:
        ans[i] = a[-1]+1
print(*ans)




