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)

#[allow(unused_imports)]
use proconio::{input, marker::*};

#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};

#[allow(unused_imports)]
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*).unwrap();
    }
}

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

    let mut c1 = 0usize;
    for &e in &x {
        if e == '1' {
            c1 += 1;
        }
    }

    let mut c = vec![0; 2];
    if c1 == 0 {
        for _ in 0..n {
            println!("1");
        }
        return;
    } else if c1 == 1 {
        let mut vec = Vec::new();
        for i in (0..n).rev() {
            if x[i] == '1' {
                vec.push(0);
            } else {
                if i == n - 1 {
                    vec.push(2);
                } else {
                    vec.push(1);
                }
            }
        }
        for e in vec.into_iter().rev() {
            println!("{}", e);
        }
        return;
    }
    c[0] = c1 - 1;
    c[1] = c1 + 1;

    let mut base = vec![0; 2];
    for i in 0..n {
        for j in 0..2 {
            if c[j] == 0 {
                continue;
            }
            base[j] = base[j] << 1 | if x[i] == '1' { 1 } else { 0 };
            base[j] %= c[j];
        }
    }

    let mut pow = vec![1 % c[0], 1 % c[1]];
    let mut vec = Vec::new();
    for i in (0..n).rev() {
        let mut r = if x[i] == '1' {
            (base[0] + c[0] - pow[0]) % c[0]
        } else {
            (base[1] + pow[1]) % c[1]
        };
        let mut count = 1usize;
        while r > 0 {
            r %= r.count_ones() as usize;
            count += 1;
        }
        vec.push(count);
        for j in 0..2 {
            pow[j] *= 2;
            pow[j] %= c[j];
        }
    }
    for e in vec.into_iter().rev() {
        println!("{}", e);
    }
}