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)

use proconio::{input, marker::Chars};

fn main() {
    input!(n: usize, mut s: Chars);
    s.reverse();
    let p = s.iter().copied().filter(|&c| c == '1').count() as u32;

    let mut dp = vec![0; n + 1];
    for i in 1..=n {
        dp[i] = 1 + dp[i % i.count_ones() as usize];
    }

    let make_pow_of_two = |m: u32| -> Vec<u32> {
        std::iter::successors(Some(1), |x| Some(x * 2 % m))
            .take(n + 1)
            .collect()
    };
    let make_full = |m: u32| -> u32 {
        let mut coeff = 1;
        let mut ans = 0;
        for x in s.iter().copied().map(|c| c == '1') {
            if x {
                ans = (ans + coeff) % m;
            }
            coeff = (coeff * 2) % m;
        }
        ans
    };
    let pow_two_mod_p_plus_one = make_pow_of_two(p + 1);
    let full_mod_p_plus_one = make_full(p + 1);
    let pow_two_mod_p_minus_one = if p == 1 {
        None
    } else {
        Some(make_pow_of_two(p - 1))
    };
    let full_mod_p_minus_one = if p == 1 { None } else { Some(make_full(p - 1)) };

    let ans: Vec<_> = (0..n)
        .map(|i| {
            if s[i] == '0' {
                1 + dp[((full_mod_p_plus_one + pow_two_mod_p_plus_one[i]) % (p + 1)) as usize]
            } else if p == 1 {
                0
            } else {
                1 + dp[((full_mod_p_minus_one.unwrap() + (p - 1)
                    - pow_two_mod_p_minus_one.as_ref().unwrap()[i])
                    % (p - 1)) as usize]
            }
        })
        .collect();

    for x in ans.iter().copied().rev() {
        println!("{}", x);
    }
}