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 
def bfs(a,f,g):
    v = [0 if i%2 == f else -1 for i in a]
    s = [(i,0) for i,v in enumerate(a) if v%2 == f]
    
    while len(s):
        ns = [];
        for ci,c in s:
            # ci,c = s.pop()
            for i in g[ci]:
                if v[i] > c + 1 or v[i] == -1:
                    v[i] = c + 1
                    ns.append((i,c + 1))
        s = ns;
    return v
    
    
def solve():
    n = int(input())
    a = list(map(int,input().split(' ')))
    g = list(map(set,[[]]*n));
    for i,v in enumerate(a):
        if i - v >= 0:
            g[i - v].add(i)
        if i + v < len(a):
            g[i + v].add(i)
    g = list(map(list,g))
    o, e = bfs(a,0,g), bfs(a,1,g)
    print(*map(sum,zip(o,e)))
        
solve()
