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 collections import deque

# inf = open('input.txt', 'r')
# reader = (line.rstrip() for line in inf)
reader = (line.rstrip() for line in sys.stdin)
input = reader.__next__

n = int(input())
a = list(map(int, input().split()))

ans = [-1] * n
ends = {0:set(), 1:set()}
for i, el in enumerate(a):
    if i - el >= 0 and (a[i-el] + el) & 1:
#         ans[el] = 1
        ends[el&1].add(i)
    elif i + el < n and (a[i+el] + el) & 1:
#         ans[el] = 1
        ends[el&1].add(i)

# print(ends)

g = [[] for _ in range(n)]
for i, el in enumerate(a):
    if i - el >= 0 and not((a[i-el] + el) & 1):
        g[i-el].append(i)
    if i + el < n and not((a[i+el] + el) & 1):
        g[i+el].append(i)

visited = [False] * n
for i in range(2):
    queue = deque((v, 1) for v in ends[i])
    while queue:
        v, d = queue.popleft()
        if not visited[v]:
            visited[v] = True
            ans[v] = d
            for to in g[v]:
                queue.append((to, d+1))

print(*ans)
# inf.close()