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
def f(i,j):
	for x in range(i+1,j):
		s.append(x)
#for _ in range(int(input())):
n=int(input())
#	n=map(int,input().split())
#	s=input().strip()
a=list(map(int,input().split()))
b=[10**6]*n
s=[]
ind=[]
if a[0]==1:b[0]=0
else:ind.append(0)
for i in range(1,n):
	if a[i]>a[i-1]:
		b[i]=a[i-1]
		f(a[i-1],a[i])
	else:ind.append(i)
ind=ind[::-1]
for x in s:
	b[ind.pop()]=x
print(*b)