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: Runtime Error)

use std::io::Read;
use std::collections::HashMap;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let x: Vec<char> = iterator.next().unwrap().chars().collect();

    let mut popcount_memo = HashMap::new();
    let mut result = String::new();
    for i in 0..n {
        let mut xi = x.clone();
        if xi[i] == '0' {
            xi[i] = '1';
        } else {
            xi[i] = '0';
        }
        let bin: String = xi.iter().collect();
        result.push_str(
            &format!(
                "{}\n",
                f(usize::from_str_radix(&bin, 2).unwrap(), &mut popcount_memo)
            )
        );
    }
    result.trim().to_string()
}

fn f(mut x: usize, popcount_memo: &mut HashMap<usize, usize>) -> usize {
    let mut ans: usize = 0;
    while x != 0 {
        x = x % popcount(x, popcount_memo);
        ans += 1;
    }
    ans
}

fn popcount(mut n: usize, memo: &mut HashMap<usize, usize>) -> usize {
    if let Some(x) = memo.get(&n) {
        return *x;
    }

    let n_original = n;

    let mut count = 0;
    while n != 0 {
        if n % 2 == 1 {
            count += 1;
        }
        n /= 2;
    }
    memo.insert(n_original, count);
    count
}