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)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

fn dfs(n: usize) -> usize {
    if n == 0 {
        return 0;
    }
    let res = n % pop_count(n);
    dfs(res) + 1
}

fn pop_count(n: usize) -> usize {
    let mut res = 0;
    let mut n = n;
    while n > 0 {
        res += n % 2;
        n /= 2;
    }
    res
}

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

    let x = x
        .into_iter()
        .map(|ch| ch.to_digit(10).unwrap() as usize)
        .collect::<Vec<_>>();
    let mut pc = 0;
    for n in 0..n {
        if x[n] == 1 {
            pc += 1;
        }
    }
    let sum_mod_plus = x.iter().copied().fold(0, |acc, x| (acc * 2 + x) % (pc + 1));
    let sum_mod_minus = x.iter().copied().fold(0, |acc, x| {
        if pc <= 1 {
            return 0;
        }
        (acc * 2 + x) % (pc - 1)
    });
    let digits_val_mod_plus = (0..n)
        .into_iter()
        .scan(1, |acc, _| {
            let res = *acc;
            *acc = (*acc * 2) % (pc + 1);
            Some(res)
        })
        .collect::<Vec<_>>();
    let digits_val_mod_minus = (0..n)
        .into_iter()
        .scan(1, |acc, _| {
            if pc <= 1 {
                return Some(0);
            }
            let res = *acc;
            *acc = (*acc * 2) % (pc - 1);
            Some(res)
        })
        .collect::<Vec<_>>();
    for i in 0..n {
        let res = if x[i] == 0 {
            (sum_mod_plus + digits_val_mod_plus[n - i - 1]) % (pc + 1)
        } else {
            if pc == 1 {
                0
            } else if digits_val_mod_minus[n - i - 1] > sum_mod_minus {
                (pc - 1) + sum_mod_minus - digits_val_mod_minus[n - i - 1]
            } else {
                sum_mod_minus - digits_val_mod_minus[n - i - 1]
            }
        };
        let res = 1 + dfs(res);
        println!("{}", res);
    }
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}