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.
from sys import stdin
import heapq

n = int(stdin.readline())

nums = [int(x) for x in stdin.readline().split()]

high = max(nums)

no = set(nums)

need = []
for x in range(high):
    if not x in no:
        need.append(x)
        
heapq.heapify(need)

out = []

for x in range(len(nums)):
    if need:
        out.append(heapq.heappop(need))
    else:
        out.append(high+1)
    if x != n-1:
        if nums[x+1] != nums[x]:
            heapq.heappush(need, nums[x])

low = 0
things = [False for x in range(high+5)]

valid = True

for i, x in enumerate(out):
    things[x] = True
    while things[low]:
        low += 1
    if low != nums[i]:
        valid = False
        break

if valid:
    print(' '.join([str(x) for x in out]))
else:
    print(-1)
    
