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;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::f64::consts::*;

#[allow(unused)]
const INF: usize = std::usize::MAX / 4;
#[allow(unused)]
const M: usize = 1000000007;

fn rec(n: usize, cache: &mut HashMap<usize, usize>) -> usize {
    if let Some(&f) = cache.get(&n) {
        f
    } else {
        let mut m = n;
        let mut one_count = 0;
        while m > 0 {
            if m % 2 == 1 {
                one_count += 1;
            }
            m /= 2;
        }
        let f = rec(n % one_count, cache);
        cache.insert(n, f + 1);
        cache[&n]
    }
}

fn main() {
    input! {
        n: usize,
        mut x: Chars,
    }
    x.reverse();
    let one_count = x.iter().filter(|&&c| c == '1').count();

    let mut s = vec![1; n];
    let mut t = vec![1; n];
    for i in 1..n {
        s[i] = s[i - 1] * 2 % (one_count + 1);
        if one_count > 1 {
            t[i] = t[i - 1] * 2 % (one_count - 1);
        }
    }

    let mut sum_p = 0;
    let mut sum_m = 0;
    for i in 0..n {
        if x[i] == '1' {
            sum_p = (sum_p + s[i]) % (one_count + 1);
            if one_count > 1 {
                sum_m = (sum_m + t[i]) % (one_count - 1);
            }
        }
    }

    let mut result = vec![];
    let mut cache = HashMap::new();
    cache.insert(0, 0);
    for i in 0..n {
        let f = if x[i] == '0' {
            let xi = (sum_p + s[i]) % (one_count + 1);
            rec(xi, &mut cache) + 1
        } else {
            let xi = if one_count > 1 {
                (sum_m + one_count - 1 - t[i]) % (one_count - 1)
            } else {
                0
            };
            rec(xi, &mut cache) + 1
        };
        result.push(f);
    }
    result.reverse();
    for &f in &result {
        println!("{}", f);
    }
}