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())
a = [0] + [int(x) for x in input().split()]
vis = set()
q = []
ans = [-1 for i in range(n+5)]
to = [ [] for i in range(n+5) ]

for i in range(1,n+1):
    if i-a[i]>=1:
        if a[i-a[i]]%2==a[i]%2:
            to[i-a[i]].append(i)
        else:
            q.append(i)
            ans[i] = 1
            vis.add(i)
    if i+a[i]<=n:
        if a[i+a[i]]%2==a[i]%2:
            to[i+a[i]].append(i)
        elif i not in vis:
            q.append(i)
            ans[i] = 1
            vis.add(i)

head = 0
while head<len(q):
    pos = q[head]
    head += 1
    for t in to[pos]:
        if t not in vis:
            ans[t] = ans[pos]+1
            vis.add(t)
            q.append(t)

print(*ans[1:n+1])