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 
from collections import defaultdict,deque
import  threading
answer=[]
def bfs(start,end,g):
	d=[float("inf")]*(len(start)+len(end)+1)

	q=deque()
	global answer
	for i in start:
		d[i]=0
		q.append(i)
	while q:
		temp=q.popleft()
		for to in g[temp]:
			if d[to]==float("inf"):
				d[to]=d[temp]+1
				q.append(to)
	for i in end:
		if d[i]!=float("inf"):
			answer[i]=d[i]
def f(a):
	global answer
	n=len(a)
	a=[None]+a
	answer=[-1]*(len(a))
	odd=[]
	even=[]
	g=defaultdict(list)
	for i in range(1,len(a)):
		if a[i]&1:
			odd.append(i)
		else:
			even.append(i)
		if a[i]+i<=n:
			g[a[i]+i].append(i)
		if i-a[i]>=1:
			g[i-a[i]].append(i)
	bfs(odd,even,g) #kaunse odd vale even ktak jaenge
	bfs(even,odd,g) #kaunse even vale odd
	return answer[1:]

a=input()
lst=list(map(int,input().strip().split()))
print(*f(lst))

