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 minp():
	return sys.stdin.readline().strip()

def mint():
	return int(minp())

def mints():
	return map(int, minp().split())

def check(a, b):
	n = len(a)
	bb = [False]*(n+1)
	c = [0]*n
	j = 0
	z = 0
	for i in a:
		if i <= n:
			bb[i] = True
		while bb[j]:
			j += 1
		if b[z] != j:
			return False
		z += 1
	return True

def solve():
	n = mint()
	a = list(mints())
	b = [None]*n
	c = [False]*(n+1)
	for i in range(1,n):
		if a[i] != a[i-1]:
			b[i] = a[i-1]
			if a[i-1] <= n:
				c[a[i-1]] = True
	if a[-1] <= n:
		c[a[-1]] = True
	if a[0] != 0:
		b[0] = 0
		c[0] = True
	#print(b)
	#print(c)
	j = 0
	for i in range(n):
		if b[i] is None:
			while c[j]:
				j += 1
			b[i] = j
			c[j] = True
	#print(b)
	#print(check(b,a))
	if check(b,a):
		print(' '.join(map(str,b)))
	else:
		print(-1)

#for i in range(mint()):
solve()
