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::*;
use proconio::marker::*;

#[fastout]
fn run() {
    input! {
        n: usize,
        s: Bytes,
    }
    let s = s.into_iter().map(|s| (s - b'0') as i32).collect::<Vec<_>>();
    let cnt = s.iter().sum::<i32>();
    let mut rem = vec![];
    for div in (cnt - 1)..=(cnt + 1) {
        if div <= 0 {
            rem.push((0, vec![]));
            continue;
        }
        let mut val = vec![0; n];
        let mut p = 1;
        let mut sum = 0;
        for (val, c) in val.iter_mut().zip(s.iter()).rev() {
            *val = p;
            sum = (c * p + sum) % div;
            p = 2 * p % div;
        }
        rem.push((sum, val));
    }
    for i in 0..n {
        let mut c = cnt;
        if s[i] == 1 {
            c -= 1;
        } else {
            c += 1;
        }
        if c == 0 {
            println!("0");
            continue;
        }
        let mut ans = 1;
        let mut val = if s[i] == 1 {
            (rem[0].0 + c - rem[0].1[i]) % c
        } else {
            (rem[2].0 + c + rem[2].1[i]) % c
        };
        while val > 0 {
            val = val % val.count_ones() as i32;
            ans += 1;
        }
        println!("{}", ans);
    }
}

fn main() {
    run();
}