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)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let n = read::<usize>();
    let x = read::<String>().chars().collect::<Vec<_>>();
    let init_count = x.iter().filter(|&&x| x == '1').count() as u64;
    let offsets = vec![-1, 1];
    let mut init = vec![0; 2];
    let mut coefs = vec![vec![0; n]; 2];
    for i in 0..2 {
        let init_count = (init_count as i64 + offsets[i]) as u64;
        if init_count == 0 && i == 0 {
            continue;
        }

        coefs[i][0] = 1;
        for j in 1..n {
            coefs[i][j] = (2 * coefs[i][j - 1]) % init_count;
        }
        for j in 0..n {
            if x[n - 1 - j] == '1' {
                init[i] += coefs[i][j];
                init[i] %= init_count;
            }
        }
    }

    for i in 0..n {
        let mut cur_x;
        if x[i] == '0' {
            cur_x = init[1];
            cur_x += coefs[1][n - 1 - i];
            cur_x %= init_count + 1;
        } else {
            cur_x = init[0];
            if init_count == 1 {
                println!("0");
                continue;
            }
            cur_x += init_count - 1 - coefs[0][n - 1 - i];
            cur_x %= init_count - 1;
        }
        let mut count = 1;
        while cur_x != 0 {
            debug!(cur_x);
            cur_x = cur_x % cur_x.count_ones() as u64;
            count += 1;
        }
        println!("{}", count);
    }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}