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.
if __name__ == '__main__':
		
	n = int(input())
	arr = list(map(int,input().rstrip().split()))
	ans = [0]*n
	val = arr[n-1] - 1
	
	s = []

	for i in range(n-1,0,-1):
		if arr[i] - arr[i-1] == 1:
			ans[i] = arr[i-1]
			val -= 1
		else:
			if(arr[i] - arr[i-1]) > 1:
				ans[i] = arr[i-1]
				for j in range(arr[i-1] + 1, arr[i]):
					s.append(arr[i]-1)
			else:
				if s:
					ans[i] = s.pop()		
				else:
					ans[i] = arr[n-1] + 1
	if arr[0] != 1: 
		ans[0] = 1
	else:
		ans[0] = 0
	print (*ans, sep = " ")