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 std::io;
use std::io::Read;

fn main(){
    let mut stdin = io::stdin();
    let mut lines_str = String::new();
    let _ = stdin.read_to_string(&mut lines_str);
    let lines_vec = lines_str.lines().collect::<Vec<&str>>();
    let n :usize = lines_vec[0].trim().parse().unwrap();
    let x_vec = lines_vec[1].trim().chars().collect::<Vec<char>>();

    // compute popcount(X)
    let mut popcount_x = 0;
    for &c in &x_vec {
        if c == '1' {
            popcount_x += 1;
        }
    }
    let popcount_x = popcount_x;

    // compute X % (popcount(X)+1) by DP
    let mut x_mod_pplus1 = 0;
    for &c in &x_vec {
        if c == '0' {
            x_mod_pplus1 = (2 * x_mod_pplus1) % (popcount_x + 1);
        } else if c == '1' {
            x_mod_pplus1 = (2 * x_mod_pplus1 + 1) % (popcount_x + 1);
        }
    }
    let x_mod_pplus1 = x_mod_pplus1;

    // compute X % (popcount(X)-1) by DP
    // only if popcount(X) is larger than 1
    let mut x_mod_pminus1 = 0;
    if popcount_x >= 2 {
        for &c in &x_vec {
            if c == '0' {
                x_mod_pminus1 = (2 * x_mod_pminus1) % (popcount_x - 1);
            } else if c == '1' {
                x_mod_pminus1 = (2 * x_mod_pminus1 + 1) % (popcount_x - 1);
            }
        }
    }
    let x_mod_pminus1 = x_mod_pminus1;
    
    for i in 0..n {
        if x_vec[i] == '0' {
            //xi is not 0
            let mut diff = 1;
            let mut j = i;
            while j != n-1 {
                diff = (2 * diff) % (popcount_x + 1);
                j += 1;
            }
            let xi_mod_pplus1 = (diff + x_mod_pplus1) % (popcount_x + 1);
            println!("{}", 1 + f(xi_mod_pplus1));
        } else if x_vec[i] == '1' {
            if popcount_x == 1 {
                println!("0");
                continue;
            }
            //xi is not 0
            let mut diff = 1;
            let mut j = i;
            while j != n-1 {
                diff = (2 * diff) % (popcount_x - 1);
                j += 1;
            }
            let xi_mod_pminus1 = (diff + x_mod_pminus1) % (popcount_x - 1);
            println!("{}", 1 + f(xi_mod_pminus1));
        }
    }
}

fn popcount(n :u64) -> u64 {
    let mut result = 0;
    let mut m = n;
    while m != 0 {
        if m & 1 != 0 {
            result = result + 1;
        }
        m = m / 2;
    }
    return result;
}

fn f(n :u64) -> u64 {
    if n == 0 {
        return 0;
    } else {
        return 1 + f(n % popcount(n));
    }
}