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: Runtime Error)

//! Framework <https://github.com/vain0x/procon>

#![allow(non_snake_case)]
#![allow(unused_imports)]

use std::collections::*;

pub struct Scan(Box<dyn Iterator<Item = &'static str>>); // '

impl Scan {
    fn new() -> Self {
        let mut buf = String::new();
        let read_line = move || {
            std::io::stdin().read_line(&mut buf).unwrap();
            Box::leak(buf.split_off(0).into_boxed_str()).split_whitespace()
        };
        Scan(Box::new(std::iter::repeat_with(read_line).flatten()))
    }

    pub fn word<T: std::str::FromStr>(&mut self) -> T {
        self.0.next().unwrap().parse().ok().unwrap()
    }

    pub fn list<T: std::str::FromStr>(&mut self, len: usize) -> Vec<T> {
        std::iter::repeat_with(|| self.word()).take(len).collect()
    }
}

use num_bigint::BigUint;
use std::mem::replace;

fn solve(X: &str) -> Vec<usize> {
    let N = X.len();

    let bits = X.bytes().map(|b| b - b'0').collect::<Vec<u8>>();
    let pc = X.bytes().filter(|&b| b != b'0').count();
    let x = BigUint::from_radix_be(&bits, 2).unwrap();

    let xp = (x.clone() % (pc + 1)).to_string().parse::<usize>().unwrap();
    let xm = if pc > 1 {
        (x.clone() % (pc - 1)).to_string().parse::<usize>().unwrap()
    } else {
        0
    };

    #[cfg(debug_assertions)]
    eprintln!(
        "\x1B[33m{}\x1B[0m = {:?}",
        "(&bits, pc, &x, &xp, &xm)",
        (&bits, pc, &x, &xp, &xm)
    );

    let mut ts = vec![];
    let mut wp = 1_usize;
    let mut wm = 1_usize;

    for i in 0..N {
        let mut y = if bits[N - 1 - i] == 0 {
            let m = pc + 1;
            (xp + wp) % m
        } else {
            let m = pc - 1;
            (xm + m - wm) % m
        };

        #[cfg(debug_assertions)]
        eprintln!("\x1B[33m{}\x1B[0m = {:?}", "(i, y)", (i, y));

        let mut t = 1;
        while y != 0 {
            let pc = usize::count_ones(y) as usize;
            y %= pc;
            t += 1;
        }

        ts.push(t);

        wp *= 2;
        wp %= pc + 1;

        if pc > 1 {
            wm *= 2;
            wm %= pc - 1;
        }
    }

    ts.reverse();
    ts
}

fn main() {
    let mut scan = Scan::new();

    let N = scan.word::<usize>();
    let X = scan.word::<String>();

    let ts = if X.bytes().all(|b| b == b'0') {
        vec![0; N]
    } else {
        solve(&X)
    };

    for t in ts {
        println!("{}", t);
    }
}