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.
maxn = 10**6+5
vis = [False for i in range(maxn)]
n = int(input())
arr = list(map(int,input().split()))
for i in arr:
	vis[i] = True
ans = []
cnt = 0
for i in range(n):
	if i!=0:
		if arr[i]==arr[i-1]:
			while(vis[cnt]):
				cnt+=1
			vis[cnt] = True
			ans.append(cnt)
		else:
			ans.append(arr[i-1])
	else:
		while(vis[cnt]):
			cnt+=1
		ans.append(cnt)
		vis[cnt] = True
print(*ans)