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 sys import stdin,stdout
from collections import deque


n = int(stdin.readline())

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

pares, impares = [],[]

ady = [[] for i in range(n)]

sol = ['-1' for i in range(n)]


for i in range(n):
    ai = a[i]
    izq = i-ai
    der = i+ai
    if izq >= 0:
        ady[izq].append(i)
    if der < n:
        ady[der].append(i)
    if ai%2 == 0:
        pares.append(i)
    else:
        impares.append(i)

def BFS(red,blue):
    d = [-1 for i in range(n)]
    q = deque()
    for u in red:
        d[u] = 0
        q.append(u)
    while len(q)>0:
        v = q.popleft()
        for i in ady[v]:
            if d[i] == -1:
                d[i] = d[v] + 1
                q.append(i)
    for i in blue:
        if d[i] != -1:
            sol[i] = str(d[i])

BFS(pares,impares)
BFS(impares,pares)
stdout.write(' '.join(sol))