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: Wrong Answer)

#![allow(unused_imports)]
#![allow(unused_macros)]
use std::cmp::{max, min};
use std::collections::*;
use std::io::{stdin, Read};

#[allow(unused_macros)]
macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

#[allow(unused_macros)]
macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, ($($t: tt),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, u1) => {
        $it.next().unwrap().parse::<usize>().unwrap() -1
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

#[cfg(debug_assertions)]
macro_rules! debug {
    ($( $args:expr ),*) => { eprintln!( $( $args ),* ); }
}

#[cfg(not(debug_assertions))]
macro_rules! debug {
    ($( $args:expr ),*) => {
        ()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(it, n: usize);
    let mut x: Vec<usize> = it
        .next()
        .unwrap()
        .chars()
        .map(|c| c.to_digit(10).unwrap() as usize)
        .collect();
    x.reverse();
    let s: usize = x.iter().sum();
    if s == 0 {
        for _ in 0..n {
            println!("{}", 1);
        }
        return;
    } else if s == 1 {
        for i in 0..n {
            if x[i] == 1 {
                println!("{}", 0);
            } else {
                println!("{}", 1);
            }
        }
        return;
    }
    let mut rm1: Vec<usize> = vec![];
    let mut cur = if s > 1 { 1 % (s - 1) } else { 0 };
    for i in 0..n {
        rm1.push(cur);
        cur = (cur * 2) % (s - 1);
    }
    let mut rp1: Vec<usize> = vec![];
    let mut cur = 1 % (s + 1);
    for _ in 0..n {
        rp1.push(cur);
        cur = (cur * 2) % (s + 1);
    }

    let mut basem1 = 0;
    let mut basep1 = 0;
    for i in 0..n {
        if x[i] == 1 {
            if s > 1 {
                basem1 = (basem1 + rm1[i]) % (s - 1);
            }
            basep1 = (basep1 + rp1[i]) % (s + 1);
        }
    }

    for i in (0..n).rev() {
        let mut r = if x[i] == 0 {
            (basep1 + rp1[i]) % (s + 1)
        } else {
            if s == 1 {
                0
            } else {
                (basem1 + (s - 1) - rm1[i]) % (s - 1)
            }
        };
        let mut count = 0;
        while r > 0 {
            r = r % (r.count_ones() as usize);
            count += 1;
        }
        println!("{}", count + 1);
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}