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

def read(): return int(input())
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()


def solve():
    n = read()
    a = get_array()
    a = [ 0 ] + a
    b = [ -1 for i in range(n + 1) ]
    used = [ False for i in range(n + 1) ]
    for i in range(1, n + 1):
        if a[i] != a[i - 1]:
            b[i] = a[i - 1]
            used[b[i]] = True
    used[a[n]] = True
    m = 0
    for i in range(1, n + 1):
        if b[i] == -1:
            while used[m]: m += 1
            b[i] = m
            used[m] = True
    for i in range(1, n + 1):
        print("%d " % b[i], end = '')
    print()

solve()