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()))

b = [-1 for k in range(n)]
existe = [False for k in range(100005)]
for k in range(n):
    if(a[k] != a[k-1] and k != 0):
        b[k] = a[k-1]
        existe[b[k]] = True
        
existe[a[n-1]] = 1
m = 0
#print(b)
for k in range(n):
    while(existe[m]):
        m += 1
    if(b[k] == -1):
        b[k] = m
        existe[m] = True
    print(b[k], end=" ")
#print(b)
