You are given an array a consisting of n integers. In one move, you can jump from the position i to the position i - a_i (if 1 ≤ i - a_i) or to the position i + a_i (if i + a_i ≤ n).

For each position i from 1 to n you want to know the minimum the number of moves required to reach any position j such that a_j has the opposite parity from a_i (i.e. if a_i is odd then a_j has to be even and vice versa).

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a.

Output

Print n integers d_1, d_2, ..., d_n, where d_i is the minimum the number of moves required to reach any position j such that a_j has the opposite parity from a_i (i.e. if a_i is odd then a_j has to be even and vice versa) or -1 if it is impossible to reach such a position.

Example

Input


10
4 5 7 6 7 5 4 4 6 4


Output


1 1 1 2 -1 1 1 3 1 1 
n=int(input())
arr=list(map(int,input().split()))

adj={}
ans=[-1]*n
queue=[]

for i in range(n):
	flg=True
	if i+arr[i]<n:
		if arr[i+arr[i]]%2==arr[i]%2:
			try:
				adj[i+arr[i]].append(i)
			except:
				adj[i+arr[i]]=[i]
		else:
			ans[i]=1
			flg=False
			queue.append(i)
	if i-arr[i]>=0:
		if arr[i-arr[i]]%2==arr[i]%2:
			try:
				adj[i-arr[i]].append(i)
			except:
				adj[i-arr[i]]=[i]
		elif flg:
			ans[i]=1
			queue.append(i)
# for i in range(n):
	# print(adj.get(i,[]))
# print(queue)
q_index=0
while q_index<len(queue):
	z=queue[q_index]
	q_index+=1
	for i in adj.get(z,[]):
		if ans[i]==-1:
			ans[i]=ans[z]+1
			queue.append(i)
		# else:
		# 	if ans[i]>ans[z]+1:
		# 		ans[i]=ans[z]+1
		# 		queue.append(i)
print(*ans)