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)

use proconio::input;
use proconio::marker::Chars;

fn main() {
    input! {
        n: usize,
        x_s: Chars,
    }

    let pop: i64 = pop_count(&x_s);
    // popが1だったら？
    if pop == 1 {
        for c in x_s {
            if c == '0' {
                println!("{}", 1);
            } else {
                println!("{}", 0);
            }
        }
        return
    }

    let x0: i64 = rem(&x_s, pop + 1);
    let x1: i64 = rem(&x_s, pop - 1);
    for i in 0..n {
        let modulo =  if x_s[i] == '0' { pop + 1 } else { pop - 1 };
        let x3 = if x_s[i] == '0' {
            let d = pow_mod(2, (n - i - 1) as i64, modulo);
            (x0 + d) % modulo
        } else {
            let d = pow_mod(2, (n - i - 1) as i64, modulo);
            ((x1 - d) % modulo + modulo) % modulo
        };
        // dbg!(x3);
        println!("{}", f(x3) + 1);
    }
}

fn f(n: i64) -> i64 {
    let mut n = n;
    let mut count = 0i64;
    while n > 0 {
        let pop = n.count_ones() as i64;
        n = n % pop;
        count += 1;
    }
    count
}

fn pow_mod(n: i64, p: i64, modulo: i64) -> i64 {
    if p == 0 {
        return 1;
    }

    if p % 2 == 0 {
        let x = pow_mod(n, p / 2, modulo);
        (x * x) % modulo
    } else {
        n * pow_mod(n, p - 1, modulo)
    }
}

fn pop_count(x_s: &Vec<char>) -> i64 {
    x_s.iter().filter(|c| **c == '1').count() as i64
}

fn rem(x_s: &Vec<char>, modulo: i64) -> i64 {
    let mut x_r = x_s.clone();
    x_r.reverse();
    let mut ret = 0i64;
    for (i, c) in x_r.iter().copied().enumerate() {
        if c == '1' {
            ret += pow_mod(2, i as i64, modulo);
            ret = ret % modulo;
        }
    }

    ret % modulo
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn pow_mod_test() {
        assert_eq!(pow_mod(2, 3, 3), 2);
        assert_eq!(pow_mod(2, 0, 3), 1);
        assert_eq!(pow_mod(3, 2, 5), 4);
    }

    #[test]
    fn pop_count_test() {
        assert_eq!(pop_count(&"111".chars().collect()), 3);
        assert_eq!(pop_count(&"101".chars().collect()), 2);
    }

    #[test]
    fn rem_test() {
        assert_eq!(rem(&"111".chars().collect(), 3), 1);
    }

    #[test]
    fn f_test() {
        assert_eq!(f(1), 1);
        assert_eq!(f(0), 0);
    }
}