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)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::iter::*;
#[allow(unused_imports)]
use std::*;

macro_rules! test {
    ($($input:expr => $output:expr),* $(,)*) => {
#[test]
        fn solve_test() {
            $(
                assert_eq!(solve($input), $output);
            )*
        }
    };
}
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

macro_rules! stdin {
    () => {{
        use std::io::Read;
        let mut s = String::new();
        std::io::stdin().read_to_string(&mut s).unwrap();
        s
    }};
}

fn main() {
    println!("{}", solve(&stdin!()));
}
fn solve(src: &str) -> String {
    input! {
        source = src,
        n:usize,
        x:chars,
    }
    let x = x
        .iter()
        .map(|&c| if c == '1' { 1 } else { 0 })
        .collect::<Vec<_>>();
    let pc: isize = x.iter().fold(0, |s, v| s + v);
    let mut ans = vec![0; n];
    for b in 0..2 {
        let npc = if b == 0 { pc + 1 } else { pc - 1 };
        if npc <= 0 {
            continue;
        }
        let mut r0 = 0;
        for xi in x.iter() {
            r0 = (r0 * 2) % npc;
            r0 += xi;
        }
        let mut k = 1;
        for (i, &xi) in x.iter().enumerate().rev() {
            if xi == b {
                let r = if b == 0 {
                    (r0 + k) % npc
                } else {
                    (r0 + npc - k) % npc
                };
                ans[i] = f(r as u32) + 1;
            }
            k = (k * 2) % npc;
        }
    }

    let mut out = String::with_capacity(n * 2);
    for a in ans.iter() {
        out.push_str(&format!("{}\n", a));
    }
    out
}

fn f(x: u32) -> u32 {
    if x > 0 {
        f(x % popcount(x)) + 1
    } else {
        0
    }
}

fn popcount(x: u32) -> u32 {
    x.count_ones()
}
test! {
    include_str!("d1_in.txt") => include_str!("d1_out.txt"),
    include_str!("d2_in.txt") => include_str!("d2_out.txt"),
    include_str!("d3_in.txt") => include_str!("d3_out.txt"),
}