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 math
def findholes(arr):
    holes = [False]*100005
    for a in arr:
        holes[a] = True
    ret = []
    for i,h in enumerate(holes):
        if h == False:
            ret.append(i)
    return ret

def findb(arr):
    holes = findholes(arr)
    holes = holes[::-1]
    b_array = []
    for i,a in enumerate(arr):
        if i>0 and arr[i]>arr[i-1]:
            holes.append(arr[i-1])
        currhole = holes.pop(-1)
        b_array.append(currhole)
        if a < 0:
            return []
        if a != 0:
            if len(holes)>0:
                if holes[-1]<=a-1:
                    return []
    return b_array
n = int(input())
a = [int(x) for x in input().split(" ")]
ans = findb(a)
if len(ans)==0:
    print(-1)
else:
    print(" ".join([str(x) for x in ans]))