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.buffer.readline
def I(): return(list(map(int,input().split())))
def sieve(n):
	a=[1]*n
	for i in range(2,n):
	    if a[i]:
	        for j in range(i*i,n,i):
	            a[j]=0
	return a

# for __ in range(int(input())):
n=int(input())
a=I()
b=[0]*n
# mins=[-1]*n
# mins[-1]=a[-1]
# for i in reversed(range(n-1)):
# 	mins[i]=min(a[i],mins[i+1])
# print(*mins)
m=max(a)
a=a+[0]
j=0
i=0
covered=-1
while(i<n):
	if covered<a[j]-1:
		b[i]=covered+1
		covered+=1
	else:
		b[i]=m+1
	if j<=i:
		while(j<n-1 and a[j]==a[j+1]):
			j+=1
		j+=1#can be n
	# print(j)
	i+=1
print(*b)


	



