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.

n=int(input())
a=list(map(int,input().split()))

def add(i):
	r=[]
	if c[i]:
		while c[i]:
			r.append(i)
			i+=1
	return(i,r)


aa=set(a)

f={i:0 for i in aa}
for i in a:f[i]+=1

c=[False for i in range(n+20)]

for i in a:
	c[i]=True

d=[i for i in aa][::-1]

rest=set()
i=0
b=[]

while d:
	if len(rest)>1:
		print(-1)
		exit()
	t=0
	x=d.pop()
	if rest:
		b.append(rest.pop())
		t+=1

	rest.add(x)
	
	while t<f[x]:
		i,r=add(i)
		b.append(i)
		i+=1
		rest.union(r)
		t+=1
print(*b)