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 
import sys
from itertools import chain, permutations

n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split(' ')))


result = [-1 for _ in a]
jumps_to = [[] for _ in a]

for i in range(len(a)):
	left = i - a[i] 
	right = i + a[i]
	if left >= 0:
		jumps_to[left].append(i)
		if (a[left] % 2) != (a[i] % 2):
			result[i] = 1
	if right < n:
		jumps_to[right].append(i)
		if (a[right] % 2) != (a[i] % 2):
			result[i] = 1

queue = []
queue_ptr = 0
for i in range(n):
	if result[i] == 1:
		queue.append(i)

while queue_ptr < len(queue):
	current = queue[queue_ptr]
	for to in jumps_to[current]:
		if result[to] == -1 or result[to] > result[current] + 1:
			queue.append(to)
			result[to] = result[current] + 1
	queue_ptr += 1


#print(jumps_to)

print(" ".join(map(str, result)))