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 Rust (1.42.0) (judge verdict: Accepted)

//use itertools::Itertools;
use std::cmp;
use std::collections::BTreeMap;
use std::collections::BTreeSet;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use std::io::Read;
use std::usize::MAX;

macro_rules! input {(source = $s:expr, $($r:tt)*) => {let mut iter = $s.split_whitespace();let mut next = || { iter.next().unwrap() };input_inner!{next, $($r)*}};($($r:tt)*) => {let stdin = std::io::stdin();let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));let mut next = move || -> String{bytes.by_ref().map(|r|r.unwrap() as char).skip_while(|c|c.is_whitespace()).take_while(|c|!c.is_whitespace()).collect()};input_inner!{next, $($r)*}};}
macro_rules! input_inner {($next:expr) => {};($next:expr, ) => {};($next:expr, $var:ident : $t:tt $($r:tt)*) => {let $var = read_value!($next, $t);input_inner!{$next $($r)*}};}
macro_rules! read_value {($next:expr, ( $($t:tt),* )) => {( $(read_value!($next, $t)),* )};($next:expr, [ $t:tt ; $len:expr ]) => {(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()};($next:expr, chars) => {read_value!($next, String).chars().collect::<Vec<char>>()};($next:expr, usize1) => {read_value!($next, usize) - 1};($next:expr, [ $t:tt ]) => {{let len = read_value!($next, usize);(0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()}};($next:expr, $t:ty) => {$next().parse::<$t>().expect("Parse error")};}

fn solve() {
    input! {
        n: usize,
        x: chars
    }

    let mut one_count = 0;
    for i in 0..n {
        if x[i] == '1' {
            one_count += 1;
        }
    }
    let one_count_plus = one_count + 1;
    let one_count_minus = one_count - 1;

    let mut plus_mods = vec![0; n];
    let mut minus_mods = vec![0; n];
    let mut base_plus = 1;
    let mut base_minus = 1;
    for i in 0..n {
        let now = n - 1 - i;
        plus_mods[now] = base_plus % one_count_plus;
        base_plus = (base_plus * 2) % one_count_plus;
        if one_count_minus != 0 {
            minus_mods[now] = base_minus % one_count_minus;
            base_minus = (base_minus * 2) % one_count_minus;
        }
    }

    let mut default_plus_mods = 0;
    let mut default_minus_mods = 0;
    for i in 0..n {
        let now = n - 1 - i;
        if x[now] == '1' {
            default_minus_mods += minus_mods[now];
            default_plus_mods += plus_mods[now];
        }
    }
    default_plus_mods %= one_count_plus;
    if one_count_minus != 0 {
        default_minus_mods %= one_count_minus;
    }

    //println!("{:?} {}", plus_mods, one_count_plus);
    //println!("{:?} {}", minus_mods, one_count_minus);

    for i in 0..n {
        let now = n - 1 - i;
        if x[i] == '1' {
            if one_count_minus != 0 {
                let mut f: u64 =
                    (default_minus_mods + one_count_minus - minus_mods[i]) % one_count_minus;
                let mut count = 1;
                while f != 0 {
                    f = f % f.count_ones() as u64;
                    count += 1;
                }
                println!("{}", count);
            } else {
                println!("{}", 0);
            }
        } else {
            let mut f: u64 = (default_plus_mods + plus_mods[i]) % one_count_plus;
            let mut count = 1;
            while f != 0 {
                f = f % f.count_ones() as u64;
                count += 1;
            }
            println!("{}", count);
        }
    }
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();

    /*
    // 入力を一括で読み込む場合
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut input = buf.split_whitespace();

    // inputに対しnext()で読み込んでいく
    let q: usize = input.next().unwrap().parse().unwrap();
    */
}

const LARGE_PRIME: u64 = 1_000_000_007;

// @〜でsnippet