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 II():
	return int(sys.stdin.readline())

def LI():
	return list(map(int, sys.stdin.readline().split()))

def MI():
	return map(int, sys.stdin.readline().split())

def SI():
	return sys.stdin.readline().strip()
n = II()
a = LI()
if a[0] not in [0,1] or a!=sorted(a):
	print(-1)
else:
	boo = True
	for i in range(n):
		if a[i]>i+1:
			boo = False
			break
	if boo == False:
		print(-1)
	else:
		b = [-1]*n
		arr = list(range(n))
		if a[-1] <n:
			arr = list(range(n+1))
			arr.pop(a[-1])
		d = {}
		for i in arr:
			d[i] = 0
		for i in range(n-1,0,-1):
			if a[i]>a[i-1]:
				b[i] = a[i-1]
				d[a[i-1]] = 1
		temp = 0
		for i in range(n):
			if b[i] == -1:
				while(d[arr[temp]] == 1):
					temp+=1
				b[i] = arr[temp]
				temp+=1
		print(*b)





