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
import math

############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input().strip()
    return(list(s[:len(s)]))
def invr():
    return(map(int,input().strip().split()))



def from_file(f):
    return f.readline


def solve(n, A):
    S = set(A)

    next_num = 0
    memory = A[0]

    b = []
    for i, ai in enumerate(A):
        # memory keeps the value in a which has just been processed
        if ai != memory:
            # check that the invariant holds:
            if memory == ai-1 or next_num >= ai:
                b.append(memory)
                memory = ai
            else:
                return None
        else:
            # keep on printing
            while next_num in S:
                next_num += 1
            b.append(next_num)
            next_num += 1
    return b




# with open('3.txt') as f:
#     input = from_file(f)
n = inp()
A = inlt()
b = solve(n, A)
if b:
    print(" ".join(map(str, b)))
else:
    print("-1")