Problem:
Score : 400 points

Problem Statement

Let \mathrm{popcount}(n) be the number of 1s in the binary representation of n.
For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.

Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)

For example, when n=7, it becomes 0 after two operations, as follows:

\mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.

\mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.

You are given an integer X with N digits in binary.
For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted.
Find f(X_1), f(X_2), \ldots, f(X_N).

Constraints

1 \leq N \leq 2 \times 10^5

X is an integer with N digits in binary, possibly with leading zeros.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print N lines. The i-th line should contain the value f(X_i).

Sample Input 1

3
011

Sample Output 1

2
1
1

X_1 = 7, which will change as follows: 7 \rightarrow 1 \rightarrow 0. Thus, f(7) = 2.

X_2 = 1, which will change as follows: 1 \rightarrow 0. Thus, f(1) = 1.

X_3 = 2, which will change as follows: 2 \rightarrow 0. Thus, f(2) = 1.

Sample Input 2

23
00110111001011011001110

Sample Output 2

2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3

Solution in OCaml (4.10.0) (judge verdict: Wrong Answer)

Scanf.scanf "%d %s" (fun n x ->
    let rec loop i acc =
        if i = n then acc else
            if x.[i] = '1' then loop (i + 1) (acc + 1)
                           else loop (i + 1) acc
    in
    let bits = loop 0 0 in
    let calc i =
        let initialmod =
            if bits = 1 && x.[i] = '1' then 0 else
            if bits = 0 && x.[i] = '0' then 0 else
                let bits = if x.[i] = '1' then bits - 1 else bits + 1 in
                let rec loop pos rest =
                    if pos = n then rest else
                        let rest = rest * 2 + (if x.[pos] = '1' then 1 else 0) in
                        let rest = if i = pos then rest lxor 1 else rest in
                        let rest = rest mod bits in
                        loop (pos + 1) rest
                in
                loop 0 0
        in
        let popcount a =
            let rec loop rest acc =
                if rest = 0 then acc else loop (rest lsr 1) (acc + (rest land 1))
            in
            loop a 0
        in
        let rec phase2 md count =
            if md = 0 then count else
                let md = md mod (popcount md) in
                phase2 md (count + 1)
        in
        phase2 initialmod 1
    in
    for i = 0 to n - 1 do
        Printf.printf "%d\n" @@ calc i
    done
)